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.math.BigInteger;

public class Main {
	public static void main(String[] args) {
		var sc = new java.util.Scanner(System.in);
		var n = sc.nextInt();
		var a = new int[n];
		var counter = new BigInteger[100001];
		for(int i=0;i<counter.length;i++) {
			counter[i]=BigInteger.ZERO;
		}
		for(int i=0;i<a.length;i++) {
			var ae = sc.nextInt();
			counter[ae]=counter[ae].add(BigInteger.ONE);
		}
		var q = sc.nextInt();
		var b = new int[q];
		var c = new int[q];
		BigInteger sum = BigInteger.ZERO;
		for(int k=0;k<counter.length;k++) {
			if(!counter[k].equals(BigInteger.ZERO)) {
				sum = sum.add(BigInteger.valueOf(k).multiply(counter[k]));
			}
		}
		for(int i=0;i<q;i++) {
			b[i]=sc.nextInt();
			c[i]=sc.nextInt();
		}
		for(int i=0;i<q;i++) {
			if(!counter[b[i]].equals(BigInteger.ZERO)) {
				sum = sum.subtract(BigInteger.valueOf(b[i]).multiply(counter[b[i]]));
				sum = sum.add(BigInteger.valueOf(c[i]).multiply(counter[b[i]]));
				counter[c[i]] = counter[c[i]].add(counter[b[i]]);
				counter[b[i]] = BigInteger.ZERO;
			}
			System.out.println(sum);
		}
	}
}