Problem: 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.

Examples

Input

4
1 2 3 4
3
1 2
3 4
2 4


Output

11
12
16


Input

4
1 1 1 1
3
1 2
2 1
3 5


Output

8
4
4


Input

2
1 2
3
1 100
2 100
100 1000


Output

102
200
2000

Solution (in JAVA):

Solution:

import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner sc = new Scanner(System.in);

		int n = sc.nextInt();
		int a[] = new int[n];

		int count[] = new int[1_00001];
		long sum = 0;
		for(int i = 0; i < n; i++){
			a[i] = sc.nextInt();
			count[a[i]] ++;
			sum += a[i];
		}

		int q = sc.nextInt();

		for(int i = 0; i < q; i++){
			int b = sc.nextInt();
			int c = sc.nextInt();

			sum += (long)(c - b)*count[b];
			count[c] += count[b];
			count[b] = 0;

			System.out.println(sum);
		}

//		long sum = 0;
//		for(int i = 0; i < 1_00001; i++){
//			sum += (long)count[i]*i;
//		}
//
//		System.out.println(sum);
	}
}