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.*;

public class Main {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);

        int N = scanner.nextInt();
        Map<Long, Long> map = new HashMap<>();

        long sum = 0;
        for (int i = 0 ; i < N ; i++) {
            Long A = scanner.nextLong();
            sum += A;
            if (map.containsKey(A)) {
                map.put(A, map.get(A) + 1);
            } else {
                map.put(A, 1L);
            }
        }

        int Q = scanner.nextInt();
        for (int i = 0 ; i < Q ; i++) {
            long B = scanner.nextLong();
            long C = scanner.nextLong();

            if (map.containsKey(B)) {
                long count = map.get(B);
                sum = sum - B * count + C * count;
                System.out.println(sum);
                map.put(B, 0L);
                if (map.containsKey(C)) {
                    map.put(C, map.get(C) + count);
                } else {
                    map.put(C, count);
                }
            } else {
                System.out.println(sum);
            }
        }
    }
}
