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.Scanner;

public class Main {
    private static final int MOD = 1_000_000_007;
    private static final String YES = "Yes";
    private static final String NO = "No";

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

        int N = sc.nextInt();
        int[] A = new int[N];
        for (int i=0; i<N; i++) {
            A[i] = sc.nextInt();
        }
        int Q = sc.nextInt();
        int[] B = new int[Q];
        int[] C = new int[Q];
        for (int i=0; i<Q; i++) {
            B[i] = sc.nextInt();
            C[i] = sc.nextInt();
        }

        long[] ans = solve(N, A, Q, B, C);
        for (int i=1; i<Q+1; i++) {
            System.out.println(ans[i]);
        }
    }

    private static long[] solve(int N, int[] A, int Q, int[] B, int[] C) {
        long[] ans = new long[Q+1];

        int[] count = new int[100000 + 1];
        for (int i=0; i<N; i++) {
            count[A[i]]++;
            ans[0] += A[i];
        }

        for (int i=0; i<Q; i++) {
            ans[i+1] = ans[i];
            ans[i+1] += (C[i] - B[i]) * count[B[i]];
            count[C[i]] += count[B[i]];
            count[B[i]] = 0;
        }

        return ans;
    }
}