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 C# (.NET Core 3.1.201) (judge verdict: Wrong Answer)

    using System;
    using System.Collections.Generic;
	using System.Linq;
    class Program
    {
      static int n;
      static int[] a = new int[100001];
      static int sum;
      static void Main()
      {
        n = int.Parse(Console.ReadLine());
        var line = Console.ReadLine();
        var _line = line.Split(" ");
        for (int i = 0; i < n; i++)
        {
          int num = int.Parse(_line[i]);
          sum += num;
          a[num]++;
        }
        int q = int.Parse(Console.ReadLine());
        for (int i = 0; i < q; i++)
        {
          line = Console.ReadLine();
          _line = line.Split(" ");
          int b = int.Parse(_line[0]);
          int c = int.Parse(_line[1]);
          sum += (c - b) * a[b];
          Console.WriteLine(sum);
          a[c] += a[b];
          a[b] = 0;
        }
      }
    }