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 sc = new Scanner(System.in);
        int n = sc.nextInt();
        long arr[]=new long[100001];
        long sum = 0;

        for(int i=0; i<n; i++) {
            int a = sc.nextInt();
            arr[a] += 1;
            sum += a;
        }

        int q = sc.nextInt();

        int[] bArr = new int[q];
        int[] cArr = new int[q];
        for(int i=0; i<q; i++) {
            bArr[i] = sc.nextInt();
            cArr[i] = sc.nextInt();
        }

        for(int i=0; i<q; i++) {
            int b = bArr[i];
            int c = cArr[i];
            long count = arr[b];
            arr[b]=0;
            arr[c]+=count;
            sum += (c-b)*count;
            System.out.print(sum);
            System.out.print("\n");
        }
    }
}