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: Accepted)

#include<iostream>
#include<vector>
#include<algorithm>
#include<cmath>
#include<ctime>
#include<queue>

using namespace std;
using ll = long long;
using vi = vector<int>;
using vvi = vector<vector<int> >;
using pii = pair<int, int>;
using ppi = pair<pii, int>;

int f_dir[2][4] = {{-1, 0, 1, 0}, {0, 1, 0, -1}}; //↑→↓←
int e_dir[2][8] = {{-1, -1, -1, 0, 0, 1, 1, 1}, {-1, 0, 1, -1, 1, -1, 0, 1}};
bool is_in_field(int y, int x, int h, int w){
  return (y >= 0 && y < h && x >= 0 && x < w);
}

int main(){
  cin.tie(0);
  ios::sync_with_stdio(false);
  int ti = clock();
  // start-----------------------------------------------
  ll n; cin >> n;
  vi a(100000, 0);
  ll s = 0;
  for(int i = 0; i < n; i++){
    int t; cin >> t;
    a[t-1]++;
    s += t;
  }
  int q; cin >> q;
  for(int i = 0; i < q; i++){
    int b, c; cin >> b >> c;
    s += (c - b) * a[b-1];
    a[c-1] += a[b-1];
    a[b-1] = 0;
    cout << s << endl;
  }

  // end-----------------------------------------------
  // cerr << 1.0 * (clock() - ti) / CLOCKS_PER_SEC << endl;
}