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 Java (OpenJDK 11.0.6) (judge verdict: Wrong Answer)

import java.util.Arrays;
import java.util.Scanner;

public class Main {

	public static void main(String[] args) {

		Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int[] a = new int[(int) (10e5+1)];
        int tmp = 0;
        long sum = 0;
        Arrays.fill(a,0);
        for (int i = 0; i < n; i++) {
            tmp = sc.nextInt();
            a[tmp] +=1;
            sum += tmp;
        }


        int q = sc.nextInt();
        long[] ans = new long[q+1];
        ans[0] = sum;
        int b, c;
    	StringBuilder sb = new StringBuilder();
        for (int i = 1; i <= q; i++) {
            b = sc.nextInt();
            c = sc.nextInt();
            sb.append(sum);
            sb.append("\n");
            ans[i] = ans[i-1] +((c-b)*a[b]);
            a[c] +=a[b];
            a[b] = 0;
        }

	}
}