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 <stdlib.h>
#include <string.h>
#include <math.h>
int acs(const void *a, const void *b) { 
  return *(int*)a - *(int*)b;
} /* 1,2,3,4.. */
int des(const void *a, const void *b) {
  return *(int*)b - *(int*)a;
} /* 8,7,6,5.. */
int cmp_char(const void *a, const void *b) { 
  return *(char*)a - *(char*)b;
} /* a,b,c,d.. */
int cmp_str(const void *a, const void *b) {
  return strcmp(*(const char **)a, *(const char **)b);
} /* aaa,aab.. */
#define min(a,b) (a < b ? a : b)
#define max(a,b) (a > b ? a : b)
#define rep(i, l, r) for (int i = l; i < r; i++)
#define MAX 100001
#define MOD 1000000007
#define INF 1000000009
typedef long long int lli;

int main(void) {
  int N, Q, A;
  int a[MAX] = {0}, B[MAX], C[MAX];
  scanf("%d", &N);
  rep(ni, 0, N) {
    scanf("%d", &A);
    a[A]++;
  }
  scanf("%d", &Q);
  rep(qi, 0, Q) scanf("%d %d", &B[qi], &C[qi]);

  int sum = 0;
  rep(ni, 0, MAX) sum += a[ni] * ni;

  rep(qi, 0, Q) {
    int add = C[qi] - B[qi];
    sum += add * a[B[qi]];  // 増加分 × 個数
    a[C[qi]] += a[B[qi]]; a[B[qi]] = 0;
    printf("%d\n", sum);
  }
  return 0;
}