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.*; 

class Main{
public static void main(String[] args){
	Scanner sc=new Scanner(System.in);
        int n=sc.nextInt();
		long[] b= new long[100001];
  		long sum=0;
  
		for(int c=0;c<n;c++){
    		int a=sc.nextInt();
            b[a]++;
          	sum+=a;
    	}
  	int q=sc.nextInt();
  
  for(int d=0;d<q;d++){
 	 int b1=sc.nextInt();
    int c=sc.nextInt();
    sum=sum-b[b1]*(long)b1+b[b1]*(long)c;
    System.out.println(sum);
    b[c]+=b[b1];
    b[b1]=0;
  
  }

    	
    }
}