cf85D Sum of Medians

In one well-known algorithm of finding the k-th order statistics we should divide all elements into groups of five consecutive elements and find the median of each five. A median is called the middle element of a sorted array (it's the third largest element for a group of five). To increase the algorithm's performance speed on a modern video card, you should be able to find a sum of medians in each five of the array.

A sum of medians of a sorted k-element set S = {a₁, a₂, ..., a_k}, where a₁ < a₂ < a₃ < ... < a_k, will be understood by as

The operator stands for taking the remainder, that is stands for the remainder of dividing x by y.

To organize exercise testing quickly calculating the sum of medians for a changing set was needed.

Input

The first line contains number n (1 ≤ n ≤ 10⁵), the number of operations performed.

Then each of n lines contains the description of one of the three operations:

• add x — add the element x to the set;
• del x — delete the element x from the set;
• sum — find the sum of medians of the set.

For any add x operation it is true that the element x is not included in the set directly before the operation.

For any del x operation it is true that the element x is included in the set directly before the operation.

All the numbers in the input are positive integers, not exceeding 10⁹.

Output

For each operation sum print on the single line the sum of medians of the current set. If the set is empty, print 0.

Please, do not use the %lld specificator to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams (also you may use the %I64d specificator).

Examples

Input

6
add 4
add 5
add 1
add 2
add 3
sum

Output

3

Input

14
add 1
add 7
add 2
add 5
sum
add 6
add 8
add 9
add 3
add 4
add 10
sum
del 1
sum

Output

5
11
13

要有一个支持自动排序的数据结构，然后支持询问所有rnk=k*5+3的数字的和

离散完实际上只有10w个不同的数

按照rnk建个线段树，每个节点存5个点，sum[i]是当前rnk%5==i的数字之和

每次加一个数，只要在rnk[a[i]]+1到n上面所有sum[k]往右移一位

#include<cstdio>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<cmath>
#include<queue>
#include<deque>
#include<set>
#include<map>
#include<ctime>
#define LL long long
#define inf 0x7ffffff
#define pa pair<int,int>
#define mkp(a,b) make_pair(a,b)
#define pi 3.1415926535897932384626433832795028841971
using namespace std;
inline LL read()
{
    LL x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
char s[10];
int n,m,x;
int d[100010];
map<int,int>mp;
struct opr{int op,x;}q[100010];
struct segtree{
    LL sum[5];
    int l,r,tag,tot;
}t[100010*8];
LL swp[5];
inline void gao(int k,int tt)
{
    int t2=0;t[k].tot+=tt;
    for (int i=0;i<5;i++)
    {
        t2=i+tt;if (t2>=5)t2-=5;if (t2<0)t2+=5;
        swp[t2]=t[k].sum[i];
    }
    for (int i=0;i<5;i++)t[k].sum[i]=swp[i];
}
inline void pushdown(int k)
{
    int tt=t[k].tag;t[k].tag=0;
    t[k<<1].tag+=tt;t[k<<1|1].tag+=tt;
    if (t[k<<1].tag>=5)t[k<<1].tag-=5;if (t[k<<1].tag<0)t[k<<1].tag+=5;
    if (t[k<<1|1].tag>=5)t[k<<1|1].tag-=5;if (t[k<<1|1].tag<0)t[k<<1|1].tag+=5;
    gao(k<<1,tt);gao(k<<1|1,tt);
}
inline void update(int k)
{
    for (int i=0;i<5;i++)t[k].sum[i]=t[k<<1].sum[i]+t[k<<1|1].sum[i];
}
inline void buildtree(int now,int l,int r)
{
    t[now].l=l;t[now].r=r;
    if (l==r)return;
    int mid=(l+r)>>1;
    buildtree(now<<1,l,mid);
    buildtree(now<<1|1,mid+1,r);
}
inline void rotate(int now,int x,int y,int d)
{
    pushdown(now);
    int l=t[now].l,r=t[now].r;
    if(l==x&&r==y)
    {
        t[now].tag+=d;
        gao(now,d);
        return;
    }
    int mid=(l+r)>>1;
    if (y<=mid)rotate(now<<1,x,y,d);
    else if (x>mid)rotate(now<<1|1,x,y,d);
    else rotate(now<<1,x,mid,d),rotate(now<<1|1,mid+1,y,d);
    update(now);
}
inline void insert(int now,int x,int z)
{
    pushdown(now);
    int l=t[now].l,r=t[now].r;
    if (l==x&&r==x)
    {
        if (z==1)t[now].sum[t[now].tot%5]+=d[x];
        else if (z==-1)t[now].sum[t[now].tot%5]-=d[x];
        return;
    }
    int mid=(l+r)>>1;
    if (x<=mid)insert(now<<1,x,z);
    else insert(now<<1|1,x,z);
    update(now);
}
int main()
{
    m=read();
    for (int i=1;i<=m;i++)
    {
        scanf("%s",s+1);
        if (s[1]=='a')
        {
            x=read();
            q[i].op=1;q[i].x=x;
            d[++n]=x;
        }else if (s[1]=='d')
        {
            x=read();
            q[i].op=2;q[i].x=x;
        }else q[i].op=3;
    }
    if (!n)
    {
        for (int i=1;i<=m;i++)puts("0");
        return 0;
    }
    sort(d+1,d+n+1);
    for (int i=1;i<=n;i++)mp[d[i]]=i;
    for (int i=1;i<=m;i++)if (q[i].x)q[i].x=mp[q[i].x];
    buildtree(1,1,n);
    for (int i=1;i<=m;i++)
    {
        if (q[i].op==1)
        {
            rotate(1,q[i].x,n,1);
            insert(1,q[i].x,1);
        }else if (q[i].op==2)
        {
            rotate(1,q[i].x,n,-1);
            insert(1,q[i].x,-1);
        }else if (q[i].op==3)
        {
            printf("%lld
",t[1].sum[3]);
        }
    }
}