Problem:
Score : 400 points

Problem Statement

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.

Sample Input 1

4
1 2 3 4
3
1 2
3 4
2 4

Sample Output 1

11
12
16

Initially, the sequence A is 1,2,3,4.

After each operation, it becomes the following:

2, 2, 3, 4

2, 2, 4, 4

4, 4, 4, 4

Sample Input 2

4
1 1 1 1
3
1 2
2 1
3 5

Sample Output 2

8
4
4

Note that the sequence A may not contain an element whose value is B_{i}.

Sample Input 3

2
1 2
3
1 100
2 100
100 1000

Sample Output 3

102
200
2000

Solution in Java (OpenJDK 11.0.6) (judge verdict: Accepted)

import java.util.Scanner;

public class Main {

	//例題は正解
	public static void main(String[] args) {
		//input
		Scanner sc = new Scanner(System.in);

		int N = sc.nextInt(); //N個の自然数

		int[] A = new int[N]; //N個の数列A
		for(int i=0; i<N; i++) {
			A[i] = sc.nextInt();
		}

		int Q = sc.nextInt(); //操作を行う回数

		int[][] BC = new int[2][Q]; //BとCを格納する配列
		for(int i=0; i<Q; i++) {
			BC[0][i] = sc.nextInt(); //Bを格納
			BC[1][i] = sc.nextInt(); //Cを格納
		}

		//メイン処理
		int[] list = new int[100001]; //メイン処理用配列
		long sum = 0; //合計

		//各数字が何回でてくるかカウント
		for(int i=0; i<N; i++) {
			list[A[i]]++;
		}

		//合計初期値
		for(int i=0; i<N; i++) {
			sum += A[i];
		}

		int r = 0;
		for(int i=0; i<Q; i++) {
			r = list[BC[1][i]]; //計算用に退避
			list[BC[1][i]] += list[BC[0][i]]; //Cに変えた分カウントアップ
			sum += ((long)BC[1][i]) * ((long)list[BC[1][i]] - (long)r); //増えた分sumに追加
			sum -= (long)BC[0][i] * (long)list[BC[0][i]]; //Bの部分は合計から減算
			list[BC[0][i]] = 0; //Bは数列から消える
			System.out.println(sum);
		}
	}

}