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 sc = new Scanner(System.in);
		
		int n = sc.nextInt();
		int[] a = new int[n];
		long[] bool = new long[100001];
		for(int i = 0 ; i < 100001; i++) {
			bool[i]	 = 0;
		}
		//System.out.println(1);
		for(int i = 0; i < n; i++) {
			a[i] = sc.nextInt();
			bool[a[i]]++;
		}
		//System.out.println(2);
		int q = sc.nextInt();
		int[] b = new int[q];
		int[] c = new int[q];
		for(int i = 0; i < q; i++) {
			b[i] = sc.nextInt();
			c[i] = sc.nextInt();
		}
		//System.out.println(3);
		long[] sa = new long[q];
		java.util.Arrays.sort(a);
		for(int i = 0; i < q; i++) {
			if(bool[b[i]] > 0) {
				//System.out.println(bool[b[i]]);
				sa[i] += bool[b[i]]*(c[i] - b[i]);
				bool[c[i]] += bool[b[i]];
				bool[b[i]] = 0;
			}
		}
		long sum = 0;
		for(int i = 0; i < n; i++) {
			sum+= (long)a[i];
		}
		for(int i = 0; i < q; i++) {
			sum+= sa[i];
			System.out.println(sum);
		}
	}
}