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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Arrays;

public class Main {

	public static void main(String[] args) throws IOException {
		// TODO Auto-generated method stub
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		int n=Integer.valueOf(br.readLine());
		long[] arr=new long[n];
		long[] counter=new long[100001];
		long sum=0;
		String[] cmd=br.readLine().split(" ");
		for(int i=0;i<n;i++)
		{
			arr[i]=Long.valueOf(cmd[i]);
			sum+=arr[i];
			counter[(int)arr[i]]++;
		}
		int q=Integer.valueOf(br.readLine());
		for(int i=0;i<q;i++)
		{
			cmd=br.readLine().split(" ");
			long x=Long.valueOf(cmd[0]);
			long y=Long.valueOf(cmd[1]);
			long diff=y-x;
			sum=sum+diff*counter[(int)x];
			counter[(int)y]=counter[(int)x]+counter[(int)y];
			counter[(int)x]=0;
			System.out.println(sum);
		}
	}

}
