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: Wrong Answer)

#include <algorithm>
#include<iostream>
#include<vector>
#include<deque>
#include<queue>
#include<stack>
#include<list>
#include<map>
#include<set>
#include<string>
#include <sstream>
#include<bitset>

#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<limits.h>
const int INF = 0x7fffffff;
using lll = long long;
using ull = unsigned long long;
using namespace std;

int box[100001];

int main(){

  lll ii,jj,kk;
  vector<int> ret;
  lll n;

  cin >> n;

  vector<lll> a(n);
  lll sum;
  
  memset(box,0,sizeof(box));
  sum =0;

  for(ii=0;ii<n;ii++){

    cin >> a[ii];
    box[a[ii]]++;
    sum += a[ii];

  }
  
  lll q;

  cin >> q;

  lll b,c;

  for(ii=0;ii<q;ii++){

    cin >> b >> c;

    sum -= b * box[b];

    box[c] += box[b];
    box[b] =  0;

    
    sum += c * box[c];



    
    //sum = 0;
    //
    //for(jj=0;jj<n;jj++){
    //
    //  if(a[jj] == b){
    //    a[jj] = c;
    //  }
    //  
    //  sum += a[jj];
    //  
    //}
    
    
    cout << sum << endl;
    
  }


  return 0;
}