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 <iomanip>
#include <string>
#include <vector>
#include <cmath>
#include <stack>
#include <set>
#include <algorithm>
#include <sstream>
#include <map>
#include <unordered_map>
using namespace std;


using vc = vector<char>;
typedef vector<int> vi;
typedef vector<string> vs;
typedef vector<vector<int>> vvi;
typedef pair<int, int> pii;

void process(const string& t) {
}

int main() {
  int t, n;
  string s;
  int x, y;
  int ans = 0;
  while (cin >> n) {
    vector<int> v(n);
    unordered_map<int, int> m;
    long long sum = 0;
    for (int i = 0; i < n; ++i) {
      cin >>v[i];
      sum += v[i];
      ++m[v[i]];
    }
    int q, x, y;
    cin >> q;
    while (q--) {
      cin >> x >> y;
      if (m.count(x) == 0) {
        cout << sum << endl;
        continue;
      }
      int cnt = m[x];
      sum += cnt * (y - x);
      cout << sum << endl;
      m.erase(x);
      m[y] += cnt;
    }
  }
  return 0;
}