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
import java.util.*;
public class Main {
	public static void main(String[] args){
		Scanner sc = new Scanner(System.in);
      	int n = sc.nextInt();
      	int[] a  = new int[n];
      	int max = 0;
      	for(int i=0;i<n;i++){
          a[i] = sc.nextInt();
        }
      	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();
        }
      	long[] tmp = new long[100001];
      	for(int i=0;i<a.length;i++){
          tmp[a[i]] += 1;
        }
      	long sum = 0;
      	for(int i=0;i<tmp.length;i++){
          sum += tmp[i]*i;
        }
      	
      	for(int i=0;i<b.length;i++){    
          sum -= tmp[b[i]]*b[i];
          sum -= tmp[c[i]]*c[i];
          tmp[c[i]] += tmp[b[i]];
          tmp[b[i]] = 0;
          sum += tmp[c[i]]*c[i];
          System.out.println(sum);
        }
	}
}
  /*
        if( (int)s.charAt(0)-48 >= 0 && (int)s.charAt(0)-48 <=48 ){
*/