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) {

		Scanner scan = new Scanner(System.in);

		int N = scan.nextInt();
		long S = 0;
		int a = 0;
		int[] num = new int[100005];
		for (int i = 0; i < N; i++) {
			a = scan.nextInt();
			num[a]++;
			S += a;
		}

		int Q = scan.nextInt();
		int b = 0;
		int c = 0;
		long[] ans = new long[Q];

		for (int i = 0; i < Q; i++) {
			b = scan.nextInt();
			c = scan.nextInt();
			S += num[b] * (c-b);
			num[c] += num[b];
			num[b] = 0;
			ans[i] = S;
		}

		for (int i = 0; i < Q; i++) {
			System.out.println(ans[i]);
		}

	}

}