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 pb push_back
#define mp make_pair
#define F first
#define S second
#define FOR(i, a, b) for(int i = a; i < b; i++)
#define sizeOf(A) (sizeof(A))/(sizeof(A[0]))
#define odd(x) ((x) % (2))
#define fastio ios_base::sync_with_stdio(false), cin.tie(0);
#define fileio freopen("template.in", "r", stdin), freopen("template.out", "w", stdout);
typedef long long ll;
const ll MOD = 1e9 + 7;

int main() {
    fastio;
    //fileio;

    ll n;
    cin >> n;
    unordered_map<ll, ll> B;
    ll sum = 0;
    FOR(i, 0, 1e5) {B[i+1] = 0;}
    FOR(i, 0, n) {
        ll x;
        cin >> x;
        B[x]++;
        sum += x;
    }
    ll Q;
    cin >> Q;
    FOR(i, 0, Q) {
        ll a, b;
        cin >> a >> b;
        ll x = B[a];
        sum += (b-a)*x;
        B[b] += x;
        cout << sum << "\n";
    }
    return 0;
}