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[]) {

        // 入力
        Scanner sc = new Scanner(System.in);
        int n = Integer.parseInt(sc.next());
        long[] a = new long[n];
        int[] count = new int[100001];
        for (int i = 0; i < n; i++) {
            int num = Integer.parseInt(sc.next());
            a[i] = num;
            count[num]++;
        }
        int q = Integer.parseInt(sc.next());
        List<String> list = new ArrayList<>();
        long sum = Arrays.stream(a).sum();
        for (int i = 0; i < q; i++) {
            int b = Integer.parseInt(sc.next());
            int c = Integer.parseInt(sc.next());
            long diff = (long) (c - b) * count[b];
            count[c] += count[b];
            count[b] = 0;
            sum += diff;
            list.add(Long.toString(sum));
        }
        sc.close();

        String result = String.join("\r\n", list);

        // 出力
        System.out.println(result);
    }
}
