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.*;

public class Main {

	public static void main(String[] args) {

		// 入力
		Scanner sc = new Scanner(System.in);
		int n = Integer.parseInt(sc.next());
		long sum = 0;
		int v[];
		v = new int[100001];
		for (int i = 0; i < n; i++) {
			int x = Integer.parseInt(sc.next());

			// 数列の総和と、各数字の登場回数を保持しておく
			sum += x;
			v[x]++;
		}

		// 各クエリ
		int q = Integer.parseInt(sc.next());
		for (int i = 0; i < q; i++) {

			// 入力
			int b = Integer.parseInt(sc.next()); // 置換前
			int c = Integer.parseInt(sc.next()); // 置換後

			// 総和の更新
			sum += (c - b) * v[b];

			// 登場回数の更新
			v[c] += v[b];
			v[b] = 0;

			// 上記方法でやることで、総和を求めるのがO(1)で済む。

			// 結果出力
			System.out.println(sum);
		}

	}
}