codeforces

C. DZY Loves Colors

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

DZY loves colors, and he enjoys painting.

On a colorful day, DZY gets a colorful ribbon, which consists of n units (they are numbered from 1 to n from left to right). The color of the i-th unit of the ribbon is i at first. It is colorful enough, but we still consider that the colorfulness of each unit is 0 at first.

DZY loves painting, we know. He takes up a paintbrush with color x and uses it to draw a line on the ribbon. In such a case some contiguous units are painted. Imagine that the color of unit i currently is y. When it is painted by this paintbrush, the color of the unit becomes x, and the colorfulness of the unit increases by |x - y|.

DZY wants to perform m operations, each operation can be one of the following:

1. Paint all the units with numbers between l and r (both inclusive) with color x.
2. Ask the sum of colorfulness of the units between l and r (both inclusive).

Can you help DZY?

Input

The first line contains two space-separated integers n, m (1 ≤ n, m ≤ 105).

Each of the next m lines begins with a integer type (1 ≤ type ≤ 2), which represents the type of this operation.

If type = 1, there will be 3 more integers l, r, x (1 ≤ l ≤ r ≤ n; 1 ≤ x ≤ 108) in this line, describing an operation 1.

If type = 2, there will be 2 more integers l, r (1 ≤ l ≤ r ≤ n) in this line, describing an operation 2.

Output

For each operation 2, print a line containing the answer — sum of colorfulness.

Examples

input

Copy

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

output

Copy

8

input

Copy

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

output

Copy

3
2
1

input

Copy

10 6
1 1 5 3
1 2 7 9
1 10 10 11
1 3 8 12
1 1 10 3
2 1 10

output

Copy

129

Note

In the first sample, the color of each unit is initially [1, 2, 3], and the colorfulness is [0, 0, 0].

After the first operation, colors become [4, 4, 3], colorfulness become [3, 2, 0].

After the second operation, colors become [4, 5, 5], colorfulness become [3, 3, 2].

So the answer to the only operation of type 2 is 8.

题目意思：给出一个序列初始值为1~n，我们能对它有两种操作，1.l~r区间颜色染成x 然后我们能得到 abs(x-col[l])的色差 2.求l~r色差和

思路：线段树的区间更新，我们维护一个颜色，如果颜色相同时才能成段更新

#include <bits/stdc++.h>
#define ll long long
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
using namespace std;
const int maxn = 100005;
ll col[maxn<<2],lazy[maxn<<2],sum[maxn<<2];
void push_up(int rt)
{
    sum[rt]=sum[rt<<1]+sum[rt<<1|1];
    col[rt]=(col[rt<<1]==col[rt<<1|1]?col[rt<<1]:0);
}
void build(int l,int r,int rt)
{
    if(l==r)
    {
        col[rt]=l;
        return ;
    }
    int m=(l+r)>>1;
    build(lson);
    build(rson);
    push_up(rt);
}
void push_down(int rt,int len)
{
    if(col[rt])
    {
        col[rt<<1]=col[rt<<1|1]=col[rt];
        sum[rt<<1]+=lazy[rt]*(ll)(len-(len>>1));
        sum[rt<<1|1]+=lazy[rt]*(ll)(len>>1);
        lazy[rt<<1]+=lazy[rt];
        lazy[rt<<1|1]+=lazy[rt];
        col[rt]=lazy[rt]=0;
    }
}
void updata(int l,int r,int rt,int L,int R,ll val)
{
    if(L<=l&&r<=R&&col[rt])
    {
        sum[rt]+=abs(col[rt]-val)*(ll)(r-l+1);
        lazy[rt]+=abs(col[rt]-val);
        col[rt]=val;
        return ;
    }
    push_down(rt,r-l+1);
    int m=(l+r)>>1;
    if(L<=m) updata(lson,L,R,val);
    if(R>m) updata(rson,L,R,val);
    push_up(rt);
}
ll query(int l,int r,int rt,int L,int R)
{
    if(L<=l&&r<=R)
    {
        return sum[rt];
    }
    push_down(rt,r-l+1);
    ll ans=0;
    int m=(l+r)>>1;
    if(L<=m) ans+=query(lson,L,R);
    if(R>m) ans+=query(rson,L,R);
    return ans;
}
int main()
{
    int n,m,fg,l,r;
    ll x;
    scanf("%d %d",&n,&m);
    build(1,n,1);
    while(m--)
    {
        scanf("%d",&fg);
        if(fg==1)
        {
            scanf("%d %d %lld",&l,&r,&x);
            updata(1,n,1,l,r,x);
        }
        else
        {
            scanf("%d %d",&l,&r);
            printf("%lld
",query(1,n,1,l,r));
        }
    }
}

PS:摸鱼怪的博客分享，欢迎感谢各路大牛的指点~