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: Time Limit Exceeded)

#include <bits/stdc++.h>
using namespace std;

int main() {
    int n;
    cin >> n;
    vector<int> A(n);
    for(int i=0;i<n;i++){
        cin >> A.at(i);
    }
    int64_t f=0;
    for(int k=0;k<n;k++){
        f += A.at(k);
    }
    int q;
    cin >> q;
    vector<vector<int>> B(q,vector<int>(2));
    for(int i=0;i<q;i++){
        for(int j=0;j<2;j++){
            cin >>B.at(i).at(j);
        }
    }
    for(int i=0;i<q;i++){
        int count;
        count=0;
        for(int j=0;j<n;j++){
            
            if(A.at(j)==B.at(i).at(0)){
                A.at(j)=B.at(i).at(1);
                count++;
            }
        }
        if(count>0){
        f = f -count*(B.at(i).at(0) -B.at(i).at(1));
        }
        cout << f << endl;
    }
}