Codeforces Round #225 (Div. 2) E. Propagating tree dfs序+-线段树

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E. Propagating tree

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Iahub likes trees very much. Recently he discovered an interesting tree named propagating tree. The tree consists of n nodes numbered from 1 to n, each node i having an initial value ai. The root of the tree is node 1.

This tree has a special property: when a value val is added to a value of node i, the value -val is added to values of all the children of node i. Note that when you add value -val to a child of node i, you also add -(-val) to all children of the child of node i and so on. Look an example explanation to understand better how it works.

This tree supports two types of queries:

• "1 x val" — val is added to the value of node x;
• "2 x" — print the current value of node x.

In order to help Iahub understand the tree better, you must answer m queries of the preceding type.

Input

The first line contains two integers n and m (1 ≤ n, m ≤ 200000). The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 1000). Each of the next n–1 lines contains two integers vi and ui (1 ≤ vi, ui ≤ n), meaning that there is an edge between nodes vi and ui.

Each of the next m lines contains a query in the format described above. It is guaranteed that the following constraints hold for all queries: 1 ≤ x ≤ n, 1 ≤ val ≤ 1000.

Output

For each query of type two (print the value of node x) you must print the answer to the query on a separate line. The queries must be answered in the order given in the input.

Examples

input

5 5
1 2 1 1 2
1 2
1 3
2 4
2 5
1 2 3
1 1 2
2 1
2 2
2 4

output

3
3
0

Note

The values of the nodes are [1, 2, 1, 1, 2] at the beginning.

Then value 3 is added to node 2. It propagates and value -3 is added to it's sons, node 4 and node 5. Then it cannot propagate any more. So the values of the nodes are [1, 5, 1,  - 2,  - 1].

Then value 2 is added to node 1. It propagates and value -2 is added to it's sons, node 2 and node 3. From node 2 it propagates again, adding value 2 to it's sons, node 4 and node 5. Node 3 has no sons, so it cannot propagate from there. The values of the nodes are [3, 3,  - 1, 0, 1].

You can see all the definitions about the tree at the following link: http://en.wikipedia.org/wiki/Tree_(graph_theory)

 

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>
#include<cstdio>
#include<cmath>
#include<string>
#include<queue>
#include<algorithm>
#include<stack>
#include<cstring>
#include<vector>
#include<list>
#include<set>
#include<map>
using namespace std;
#define ll long long
#define pi (4*atan(1.0))
#define eps 1e-14
#define bug(x)  cout<<"bug"<<x<<endl;
const int N=2e5+10,M=4e6+10,inf=2147483647;
const ll INF=1e18+10,mod=1e9+7;
///   数组大小
struct is
{
    int v,next;
}edge[N<<1];
int head[N],edg,a[N];
void init()
{
    memset(head,-1,sizeof(head));
    edg=0;
}
void add(int u,int v)
{
    edg++;
    edge[edg].v=v;
    edge[edg].next=head[u];
    head[u]=edg;
}
int in[N],out[N],tot,deep[N];
void dfs(int u,int fa,int dep)
{
    in[u]=++tot;
    deep[u]=dep;
    for(int i=head[u];i!=-1;i=edge[i].next)
    {
        int v=edge[i].v;
        if(v==fa)continue;
        dfs(v,u,dep+1);
    }
    out[u]=tot;
}
/// 线段树
struct SGT
{
    int TL[N<<2],TR[N<<2],ans;
    void build(int l,int r,int pos)
    {
        TL[pos]=TR[pos]=0;
        if(l==r)return;
        int mid=(l+r)>>1;
        build(l,mid,pos<<1);
        build(mid+1,r,pos<<1|1);
    }
    void update(int L,int R,int c,int dep,int l,int r,int pos)
    {
        if(L<=l&&r<=R)
        {
            if(dep&1)
            {
                TL[pos]+=c;
                TR[pos]-=c;
            }
            else
            {
                TL[pos]-=c;
                TR[pos]+=c;
            }
            return;
        }
        int mid=(l+r)>>1;
        if(L<=mid)
            update(L,R,c,dep,l,mid,pos<<1);
        if(R>mid)
            update(L,R,c,dep,mid+1,r,pos<<1|1);
    }
    void query(int p,int dep,int l,int r,int pos)
    {
        if(dep&1)
            ans+=TL[pos];
        else
            ans+=TR[pos];
        if(l==r)return;
        int mid=(l+r)>>1;
        if(p<=mid)
            query(p,dep,l,mid,pos<<1);
        else
            query(p,dep,mid+1,r,pos<<1|1);
    }
};
SGT tree;
int main()
{
    init();
    int n,q;
    scanf("%d%d",&n,&q);
    for(int i=1;i<=n;i++)
        scanf("%d",&a[i]);
    for(int i=1;i<n;i++)
    {
        int u,v;
        scanf("%d%d",&u,&v);
        add(u,v),add(v,u);
    }
    dfs(1,-1,1);
    tree.build(1,n,1);
    while(q--)
    {
        int t;
        scanf("%d",&t);
        if(t==1)
        {
            int x,c;
            scanf("%d%d",&x,&c);
            tree.update(in[x],out[x],c,deep[x],1,n,1);
        }
        else
        {
            int x;
            scanf("%d",&x);
            tree.ans=0;
            tree.query(in[x],deep[x],1,n,1);
            printf("%d
",tree.ans+a[x]);
        }
        /*for(int i=1;i<=n;i++)
        {
            tree.ans=0;
            tree.query(in[i],deep[i],1,n,1);
            printf("%d ",tree.ans+a[i]);
        }
        printf("
");*/
    }
    return 0;
}

E. Propagating tree

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Iahub likes trees very much. Recently he discovered an interesting tree named propagating tree. The tree consists of n nodes numbered from 1 to n, each node i having an initial value ai. The root of the tree is node 1.

This tree has a special property: when a value val is added to a value of node i, the value -val is added to values of all the children of node i. Note that when you add value -val to a child of node i, you also add -(-val) to all children of the child of node i and so on. Look an example explanation to understand better how it works.

This tree supports two types of queries:

• "1 x val" — val is added to the value of node x;
• "2 x" — print the current value of node x.

In order to help Iahub understand the tree better, you must answer m queries of the preceding type.

Input

The first line contains two integers n and m (1 ≤ n, m ≤ 200000). The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 1000). Each of the next n–1 lines contains two integers vi and ui (1 ≤ vi, ui ≤ n), meaning that there is an edge between nodes vi and ui.

Each of the next m lines contains a query in the format described above. It is guaranteed that the following constraints hold for all queries: 1 ≤ x ≤ n, 1 ≤ val ≤ 1000.

Output

For each query of type two (print the value of node x) you must print the answer to the query on a separate line. The queries must be answered in the order given in the input.

Examples

input

5 5
1 2 1 1 2
1 2
1 3
2 4
2 5
1 2 3
1 1 2
2 1
2 2
2 4

output

3
3
0

Note

The values of the nodes are [1, 2, 1, 1, 2] at the beginning.

Then value 3 is added to node 2. It propagates and value -3 is added to it's sons, node 4 and node 5. Then it cannot propagate any more. So the values of the nodes are [1, 5, 1,  - 2,  - 1].

Then value 2 is added to node 1. It propagates and value -2 is added to it's sons, node 2 and node 3. From node 2 it propagates again, adding value 2 to it's sons, node 4 and node 5. Node 3 has no sons, so it cannot propagate from there. The values of the nodes are [3, 3,  - 1, 0, 1].

You can see all the definitions about the tree at the following link: http://en.wikipedia.org/wiki/Tree_(graph_theory)