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();
	    	max=Math.max(max, a[i]);
	    }
	    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();
	    	max=Math.max(max, b[i]);
	    	max=Math.max(max, c[i]);
	    }
	    long sum=0;
	    int[] count =new int[max];
	    for(int i=0;i<n;i++) {
	    	count[a[i]-1]++;
	    	sum+=a[i];
	    }
	    for(int i=0;i<q;i++) {
	    	if(b[i]>max | c[i]>max)continue;
	    	int num=count[b[i]-1];
	    	count[b[i]-1]=0;
	    	count[c[i]-1]+=num;
	    	sum+=(c[i]-b[i])*num;
	    	System.out.println(sum);
	    }
	    
	}
}
