Problem:
Score : 400 points

Problem Statement

You have a sequence A composed of N positive integers: A_{1}, A_{2}, \cdots, A_{N}.

You will now successively do the following Q operations:

In the i-th operation, you replace every element whose value is B_{i} with C_{i}.

For each i (1 \leq i \leq Q), find S_{i}: the sum of all elements in A just after the i-th operation.

Constraints

All values in input are integers.

1 \leq N, Q, A_{i}, B_{i}, C_{i} \leq 10^{5}

B_{i} \neq C_{i}

Input

Input is given from Standard Input in the following format:

N
A_{1} A_{2} \cdots A_{N}
Q
B_{1} C_{1}
B_{2} C_{2}
\vdots
B_{Q} C_{Q}

Output

Print Q integers S_{i} to Standard Output in the following format:

S_{1}
S_{2}
\vdots
S_{Q}

Note that S_{i} may not fit into a 32-bit integer.

Sample Input 1

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

Sample Output 1

11
12
16

Initially, the sequence A is 1,2,3,4.

After each operation, it becomes the following:

2, 2, 3, 4

2, 2, 4, 4

4, 4, 4, 4

Sample Input 2

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

Sample Output 2

8
4
4

Note that the sequence A may not contain an element whose value is B_{i}.

Sample Input 3

2
1 2
3
1 100
2 100
100 1000

Sample Output 3

102
200
2000

Solution in C++ (GCC 9.2.1) (judge verdict: Runtime Error)

#include<bits/stdc++.h>
using namespace std;
#define pb push_back
#define pf push_front
#define ll long long

int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    
    ll x,n,q,sum=0,a,b;
    cin>>n;
    map<int,int> mp;
    map<int,int>::iterator it,it2;

    for(int i=1; i<=n; i++){
        cin>>x;
        mp[x]++;
    }
    for(it = mp.begin(); it!=mp.end(); it++){
        sum += (it->first * it->second);
    }
    cin>>q;
    while(q--){
        cin>>a>>b;
        it = mp.find(a);
        sum -= (it->first * it->second);
        int temp = it->second;
        it2 = mp.find(b);
        if(it2==mp.end()){
            mp[b] = temp;
            sum += (b*temp);
        }
        else{
            sum -= (it2->first * it2->second);
            int temp2 = it2->second;
            mp[b] = temp+temp2;
            sum += (b*(temp+temp2));
        }
        mp.erase(it);
        cout << sum << "\n";
    }
    
    return 0;
}