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 {

	static Map<Long, Integer> map = new TreeMap<Long, Integer>();

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int N = sc.nextInt();
		long sum = 0;
		for (int i = 0; i < N; i++) {
			long a = sc.nextLong();
			if (!map.containsKey(a)) {
				map.put(a, 1);
			} else {
				map.put(a, map.get(a) + 1);
			}
			sum += a; // 初期の合計を計算
		}
		int Q = sc.nextInt();
		for (int i = 0; i < Q; i++) {
			long b = sc.nextLong();
			long c = sc.nextLong();
			// Biが含まれない場合はスキップ
			if (!map.containsKey(b)) {
				System.out.println(sum);
				continue;
			}
			// BiをCiに置き換える
			int cnt = map.get(b);
			// sumの更新
			sum += (c - b) * (long) cnt;
			// Map操作
			map.remove(b);
			if (!map.containsKey(c)) {
				map.put(c, cnt);
			} else {
				map.put(c, map.get(c) + cnt);
			}
			System.out.println(sum);
		}
		sc.close();
	}
}
