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.*;

public class Main {
	public static void main(String[] args) {
		
		Scanner keyboard = new Scanner(System.in); 
		
		int N = keyboard.nextInt();
		int[] A = new int[100010];
		long ans = 0;
		
		for(int n = 0; n<N; n++) {
			int a = keyboard.nextInt();
			A[a]++;
			ans += a;
		}
		
		int Q = keyboard.nextInt();
		StringBuilder ANS = new StringBuilder();
		
		for(int q = 0; q<Q; q++) {
			int b = keyboard.nextInt();
			int c = keyboard.nextInt();
			//掛け算してLongになるとき、先頭にLongの形式で1をかけておくと、型がlongになる
			ans += 1L*c*A[b];
			ans -= 1L*b*A[b];
			
			A[c] += A[b];
			A[b] = 0;
			
			ANS.append(Long.toString(ans) + "\n");
		}
		
		System.out.println(ANS);
		keyboard.close();
	}
}