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 <string>
#include <bits/stdc++.h>
#include <ctime>

#define FAST ios_base::sync_with_stdio(false); cin.tie(NULL)
#define TT int test_cases; cin >> test_cases; while(test_cases--)
#define TT1 int test_cases=1; while(test_cases--)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
 
using namespace std;
typedef long long int ll;

 
int main() {
    
    FAST;
    TT1{
        
        int n;cin>>n;
        vector<ll> a(n);
        map<ll,ll> m;
        ll sum=0;
        for(int i=0;i<n;i++)
        {
            cin>>a[i];
            m[a[i]]++;
            sum+=a[i];
            
        }
        int q;cin>>q;
        for(int i=0;i<q;i++)
        {
            ll b,c;cin>>b>>c;
            ll t=m[b];
            sum=sum+t*(c-b);
            m.erase(b);
            m[c]=m[c]+t;
            cout<<sum<<"\n";
        }
        
        
    }
}