/*
Time complexity: O(M * sqrt(N) * log(N))
Use the property A%B = A - (A/B)*B
Processing 3td type query can be divided in 2 parts: processing big groups 
with equal Y/i( Lj<=i<=Rj ), O(sqrt(N)) the biggest groups. One need to use segment tree or fenwick tree to track and get 
number of families on intervals. The smaller groups are processed by bruteforce.
*/

#include <cstdio>
#include <algorithm>
#include <vector>
#include <iostream>
#include <string>
#include <string.h>
#include <cmath>
#include <iomanip>
#include <set>
#include <map>
#include <queue>
 
#define X first
#define Y second
 
using namespace std;
 
const int maxn = 100100;
long long r[maxn], a[maxn];
int n, m;
void modify(int x, long long y)
{
    for (; x < maxn; x = (x|(x+1)))
        r[x] += y;
}
long long sum(int x)
{
    long long res = 0;
    for (; x >= 0; x = ((x&(x+1)) - 1))
        res += r[x];
    return res;
}
 
long long calc(long long x)
{	
    long long res=0;
    int minst = x+1;
    for (int i=1; i <= 110; i++)
	{
        int st, fin=minst-1;
        st=(x/(i+1))+1;
        if (fin==0) return res;
        res+=(sum(fin) - sum(max(st-1, 0)) )*(long long)i;
        minst=st;
    }
    for (int i=1; i<minst; i++)
        res+=(x/i)*a[i];
    return res;
}
long long gr=0;
int main ()
{
    cin>>m;
    for (int i=1; i<=m; i++)
    {
        long long aa, bb;
        cin>>aa>>bb;
	gr += bb;
        a[aa] += bb*aa;
        modify(aa, bb*aa);
    }
    cin>>n;
    for (int i=1; i<=n; i++)
    {
        char ss[3];
        cin>>ss;
	if ( ss[0] == '+' )
        {
            int x;
            cin>>x;
	    modify(x, x);
            a[x]+=x;
            gr++;
        }
        if ( ss[0] == '-' )
        {
            int x;
            cin>>x;
	    modify(x, -x);
            a[x]-=x;
            gr--;
        }
        if ( ss[0] == '?' )
        {
            long long x;
	    cin>>x;
	    long long S = gr*x, T;
            T = calc(x);
            cout<< S-T <<endl;
	}
    }
    return 0;
}