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; 
typedef long long ll;
ll mod = 1e9+7;

int main() 
{ 
/*	ll n;
	cin>>n;
	ll s = (25*n)/26;
	ll x = log(s+1)/log(26);
	ll y = (26*(pow(26,x)-1))/25;
	ll z = n - y;
	cout<<z;
	if(y==n) {
		for(int i=0; i<x; i++)
			cout<<"z";
	}
	// final digits = x+1;
	else {
		while() {
			ll r = z/26;
			ll c = z - pow(26,r);
		}
	}
*/
	int n;
	cin>>n;
	int a[n];
	map<int,int> mp;
	int sum = 0;
	for(int i=0; i<n; i++)  {
		cin>>a[i];
		mp[a[i]]= mp[a[i]]+1;
		sum += a[i];
	}

	int q;
	cin>>q;
	for(int i=0; i<q; i++) {
		int b,c;
		cin>>b>>c;
		sum -= mp[b]*b;
		sum += mp[b]*c;
		mp[c] += mp[b];
		mp[b] = 0;
		cout<<sum<<"\n";
	}

    return 0; 
}