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<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<math.h>

#define mod 1000000007
#define limit 200001
typedef long long ll;
typedef long double ld;
#define rep(i,n) for(i = 0; i < n; i++)
#define max(a,b) {if(a>b){return a;}return b;}
#define min(a,b) {if(a<b){return a;}return b;}
int up_int(const void *a, const void *b){return *(int*)a - *(int*)b;}
int down_int(const void *a, const void *b){return *(int*)b - *(int*)a;}
int up_char(const void *left, const void *right) {return strcmp((char *)left,(char *)right);}
int down_char(const void *left, const void *right) {return strcmp((char *)right,(char *)left);}

int main(){
    ll ans = 0;
    int flag = 0;
    int i,j,n,q,a[200000],b[200000],c[200000];
    ll times[200000];
    
    scanf("%d",&n);
    for(i = 0; i < n; i++) scanf("%d",&a[i]);
    scanf("%d",&q);
    for(i = 0; i < q; i++) scanf("%d %d",&b[i],&c[i]);
    
    for(i = 0; i < n; i++) times[a[i]]++; // 個数を数える
//    for(i = 1; i <= n; i++) printf("times %d ... %d\n",i,times[i]);
    for(i = 1; i <= n; i++) ans += (times[i] * i);
    
    for(i = 0; i < q; i++){
        if(times[b[i]] * (c[i] - b[i]) != 0){
            ans += (times[b[i]] * c[i]);
            ans -= (times[b[i]] * b[i]);
            times[c[i]] += times[b[i]];
            times[b[i]] = 0;
            printf("%lld\n",ans);
        }else{
            printf("%lld\n",ans);
        }
    }
    // for(i = 1; i <= n; i++) ans += (times[i] * i);
    // for(i = 1; i <= n; i++) printf("times %d ... %d\n",i,times[i]);
    // printf("%lld\n",ans);
    return 0;
}