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.HashMap;
import java.util.Map;
import java.util.Scanner;

public class Main {

    public static void main(String[] args) {
	// write your code here
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();

        Map<Integer, Integer> mp = new HashMap<>();
        long sumA = 0;
        for (int i = 0; i < n; i++) {
            int a = sc.nextInt();
            sumA += a;
            if (mp.containsKey(a)) {
                mp.put(a, mp.get(a) + 1);
            } else {
                mp.put(a, 1);
            }
        }

        int q = sc.nextInt();
        for (int i = 0; i < q; i++) {
            int b = sc.nextInt();
            int c = sc.nextInt();
            int diff = c - b;
            if (mp.containsKey(b)) {
                int t = mp.get(b);
                mp.replace(b, 0);
                if (mp.containsKey(c)) {
                    mp.replace(c, mp.get(c) + t);
                } else {
                    mp.put(c, t);
                }
                sumA += diff*t;
            }

            System.out.println(sumA);
        }
    }
    
}