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 <bits/stdc++.h>
using namespace std;
#define all(x) (x).begin(),(x).end()
#define ll long long
#define vi vector<int>
#define pii pair<int,int>
#define F first
#define S second
#define pb push_back
#define forn(b) for(int i = 0; i < b; i++)
#define rforn(b) for(int i = b - 1 ;i >= 0; i--)
#define It(m) for(auto it = m.begin(); it != m.end(); it++)
#define N 500500
#define PI 3.14159265358979323846264338327950L
#define speed ios_base :: sync_with_stdio(0);cin.tie(0);cout.tie(0);
    
const ll mod = (ll)1e9 + 7;
    
int main(){
    speed;
    int n; cin >> n;
    vi v(n);
    unsigned ll sum = 0;
    map<ll, ll> m;
    forn(n) { cin >> v[i]; sum += v[i]; m[v[i]]++; };
    int t; cin >> t;
    while(t--) {
        int b, c; cin >> b >> c;
        if(m[b] != 0) {
            int k = m[b];
            sum -= (k * b);
            sum += (k * c);
            m[b] = 0;
            m[c] += k;
        }
        cout << sum << endl;
    }   
    return 0;
}