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.*;
import java.util.Arrays;
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 sum =0;
      long []count = new long [100001];
      for(int i=0;i<n;i++){
        a[i] = sc.nextInt();
        count[a[i]]++;
        sum += a[i];
      }
      
      int q = sc.nextInt();
      long []ans = new long[q];
  
      for(int i=0;i<q;i++){
        int b = sc.nextInt();
        int c = sc.nextInt();
        
        sum -= count[b]*b;
        sum += count[b]*c;
        ans[i] = sum;
        count[c] += count[b];
        count[b] =0;
        
      }
      for(int i=0;i<q;i++){
        System.out.println(ans[i]);
      }



     

      
    }
}