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>
#define int long long
#define quickie ios_base::sync_with_stdio(false); cin.tie(NULL);
#define rep(i, a, b) for(int i=a; i<b; i++)
#define rep1(i, a, b) for(int i=a; i<=b; i++)
#define repp(i, a, b) for(int i=b-1; i>=a; i--)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define bn begin()
#define en end()
#define SZ(x) ((int)(x).size())
#define db double
#define mi map<int, int>
#define vi vector<int>
#define qi queue<int>
#define MI(x) power(x, mod-2)
#define test int t; cin >> t;
#define mod 1000000007
#define pi 3.141592653589
using namespace std;

int power(int x, int y) ;
int gcd(int a, int b) ;


signed main() {
    quickie
    int n ;
    cin >> n ;
    int a[n] ;
    int s = 0 ;
    mi m ;
    rep(i, 0, n) {
        cin >> a[i] ;
        s += a[i] ;
        m[a[i]]++ ;
    }
    int q ;
    cin >> q ;
    while(q--) {
        int b, c; 
        cin >> b >> c ;
        int xx = m[b] ;
        m[c] += xx ;
        int yy = xx*b ;
        int zz = xx*c ;
        s -= yy ;
        s += zz ;
        cout << s << "\n" ;
        m[b] = 0 ;
    }
}

int power(int x, int y) {
    int res = 1; x %= mod;
    while (y > 0) {
        if (y & 1)
            res = (res*x) % mod;
        y = y>>1;
        x = (x*x) % mod;
    }
    return res%mod;
}

int gcd(int a,int b){
    if(a==0) return b;
    return gcd(b%a,a);
}