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>

int main(void) {
	int n;
	scanf("%d",&n);
	int a[n];
	for(int i=0;i<n;i++)scanf("%d",&a[i]);
	//aの和と数値ごとの出現回数を格納する。
	int sum1=0,a_count[100000];
	for(int i=0;i<100000;i++)a_count[i]=0;//初期化
	for(int i=0;i<n;i++){
		a_count[a[i]]+=1;
	}//数値ごとの出現回数
	for(int i=0;i<n;i++)sum1+=a[i];//aの和
	int q;
	scanf("%d",&q);
	int b[q],c[q];
	for(int i=0;i<q;i++)scanf("%d%d",&b[i],&c[i]);

	for(int i=0;i<q;i++){//q回分入れ替えを行う
		int henka=0;
		//(c-b)*a_count[bi]
		henka = (c[i]-b[i])*a_count[b[i]];
		a_count[c[i]]+= a_count[b[i]];
		a_count[b[i]] = 0;
		//printf("henka:%d\n",henka);
		//入れ替え回数
		sum1+=henka;
		printf("%d\n",sum1);
	}
}