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.lang.*;
import java.io.*;

class Main {
    
    private static BufferedReader br;
    private static StringTokenizer strz; 
    
	public static void main (String[] args) throws java.lang.Exception {
		br=  new BufferedReader(new InputStreamReader(System.in));
		int n = Integer.parseInt(br.readLine());
		strz = new StringTokenizer(br.readLine());
		
		long sum = 0;
		long[] count = new long[100001];
		while(strz.hasMoreTokens()){
		    int x = Integer.parseInt(strz.nextToken());
		    sum += x;
		    ++count[x];
		}
		int q = Integer.parseInt(br.readLine());
		
		while(q-- > 0){
		    strz = new StringTokenizer(br.readLine());
		    int b = Integer.parseInt(strz.nextToken());
		    int c = Integer.parseInt(strz.nextToken());
		    
		    sum -= count[b]*b;
		    count[c] += count[b];
		    sum += count[b]*c;
		    count[b] = 0;
		    
		    System.out.println(sum);
		}
	}
}
