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<bits/stdc++.h>
using namespace std;
#define optimizar_io ios_base::sync_with_stdio(0); cin.tie(0);
#define fore(i, a, b) for(int i = a; i < b; i++)
#define fori(i, a, b) for(int i = a; i <= b; i++)
#define pb push_back
#define popb pop_back
#define mk make_pair
#define fi first
#define se second
#define endl '\n'
#define MAX 100005
typedef long long int lli;
typedef long double ld;
const double PI = acos(-1.0);
ld eps = 1e-9;

int main() {
	optimizar_io
	int n; cin >> n;
	lli b, c; 
	vector<lli> v(n);
	vector<int> bucket(MAX);
	fore(i, 0, n) {
		cin >> v[i];
		bucket[v[i]]++;
	}
	lli sumTotal = 0;
	fore(i, 0, bucket.size()) sumTotal += (bucket[i] * i);
	
	int q; cin >> q;
	fore(i, 0, q) {
		cin >> b >> c;
		sumTotal -= (b * bucket[b]);
		sumTotal -= (c * bucket[c]);
		bucket[c] += bucket[b];
		bucket[b] = 0;
		sumTotal += (bucket[c] * c);
		cout << sumTotal << endl;
	}
}