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 1.8.0) (judge verdict: Accepted)

import java.util.*;
import java.io.*;
public class Main {

	public static void main(String[] args) throws Exception{
		int num = 998244353;

		// TODO Auto-generated method stub
 		BufferedReader bf = new BufferedReader(new InputStreamReader(System.in));
 		PrintWriter out = new PrintWriter(System.out);
 		int n = Integer.parseInt(bf.readLine());
 		StringTokenizer st = new StringTokenizer(bf.readLine());
 		int[] freq = new int[100001];
 		long sum = 0;
 		for(int i = 0;i<n;i++){
 			int a= Integer.parseInt(st.nextToken());
 			freq[a]++;
 			sum+=(long)(a);
 		} 		
 		int q = Integer.parseInt(bf.readLine());
 		
 		for(int i =0 ;i<q;i++){
 			StringTokenizer st1 = new StringTokenizer(bf.readLine());
 			int a = Integer.parseInt(st1.nextToken());
 			int b = Integer.parseInt(st1.nextToken());

 				sum = sum-(long)(a)*(long)(freq[a])+(long)(b)*(long)(freq[a]);
 				out.println(sum);
 				freq[b]+=freq[a];
 				freq[a] = 0;

 		}
 		out.close();
 		
 	}
}