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;
typedef long double LD;
#define maxn 100005
#define Mod 1000000007
typedef pair<int,int> PII;
typedef map<int,int> MII;
typedef vector<int> VI;
typedef vector<vector<int>> VII;
typedef vector<long long> VLL; 
#define For(i,a,b) for(int i = (a); i < (b); ++i)
#define Rep(i,b,a) for(int i = (b); i >= (a); --i)
#define all(a) (a).begin(), (a).end()
#define __builtin_popcountll bitCountLL
#define __builtin_popcount bitCount
//__builtin_ctz 末尾0的个数, __builtin_clz 前导0的个数
//__builtin_ffs 最后一个1的位置
const int INF = 2e9;
const LL ML = 4e18;

void solve() {
    int n;
    cin >> n;
    VI a(n);
    map<int,int> mp;
    LL sum = 0;
    For(i,0,n) {
        cin >> a[i];
        mp[a[i]]++;
        sum += a[i];
    }
    int q;
    cin >> q;
    For(i,0,q) {
        int a,b;
        cin >> a >> b;
        sum += mp[a] * (b - a);
        mp[b] += mp[a];
        cout << sum << endl;
    }
}

int main() {
    ios_base::sync_with_stdio(0); cin.tie(0);
	//cout.precision(30); cout << fixed;
	//freopen("D:\\data_in.txt", "r", stdin);
    solve();
    
    /*clock_t start,finish;
    start = clock();
   
    finish = clock();
    cout << "\n" << (double)(finish - start) / (double)(CLOCKS_PER_SEC) * 1000 << "ms" << endl;
    */
    return 0;
}

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