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)

#define _USE_MATH_DEFINES
#include<iostream>
#include<string>
#include<vector>
#include<algorithm>
#include<stdio.h>
#include<cmath>
#include<stack>
#include<set>
#include<map>
#include<stdio.h>
#include<functional>
#include<queue>

using ll = long long;

using namespace std;

//ll mod = 1000000007;
ll mod = 998244353;

ll n_pow(ll a, ll n) {
	if (n == 0)return 1;
	ll t = n_pow(a, n / 2);
	t *= t;
	//t %= mod;
	if (n % 2 == 1)t *= a;
	return t /*% mod*/;
}

const int INF = 1001001001;
//const ll INF = 1e18;

int dx[] = { 1,0,-1,0 };
int dy[] = { 0,1,0,-1 };

ll cal_gcd(ll a, ll b) {
	ll m = max(a, b);
	ll n = min(a, b);
	while (m%n != 0) {
		ll b = m % n;
		m = n;
		n = b;
	}
	return n;
}

vector<pair<ll, ll>> prime_factorize(ll N) {
	vector<pair<ll,ll>> res;
	for (long long a = 2; a * a <= N; ++a) {
		if (N % a != 0) continue;
		long long ex = 0; 
		while (N % a == 0) {
			++ex;
			N /= a;
		}
		res.push_back({ a, ex });
	}
	if (N != 1) res.push_back({ N, 1 });
	return res;
}


int main() {
	int n;
	cin >> n;
	vector<int>num(100001);
	ll sum = 0;
	for (int i = 0; i < n; i++) {
		int a;
		cin >> a;
		num[a]++;
		sum += a;
	}
	int q;
	cin >> q;
	vector<int>ans(q);
	for (int i = 0; i < q; i++) {
		int b, c;
		cin >> b >> c;
		sum -= num[b] * b;
		sum += num[b] * c;
		num[c] += num[b];
		num[b] = 0;
		ans[i] = sum;
	}
	for (int i = 0; i < q; i++) {
		cout << ans[i] << endl;
	}
	return 0;
}