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[100001];
        for(int i=0;i<n;i++){
            a[sc.nextInt()]++;
        }
        int q = sc.nextInt();
        int b,c;
        long bef = sum(a);
        long ans[] = new long[q];
        for(int i=0;i<q;i++){
            b = sc.nextInt();
            c = sc.nextInt();
            ans[i] = bef+ (c-b)*a[b];
            bef = ans[i];
            a[c] += a[b];
            a[b]=0;
        }
        for (long x : ans) {
            System.out.println(x);
        }
    }

    static long sum(int[] a){
        long sum = 0;
        for (int i=0;i<a.length;i++){
            if(a[i]!=0){
                sum += a[i]*(i);
            }
        }
        return sum;
    }
}