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.*;
import java.io.*;
class Main{
  public static void main(String args[]){
    int n=readI();
    int a[]=new int[100001];
    long sum=0;
    for(int i=0;i<n;i++){
      int k=readI();
      a[k]++;
      sum+=k;
    }
    int q=readI();
    for(int ii=0;ii<q;ii++){
      int b=readI(),c=readI();
      if(a[b]!=0){
        sum+=(c-b)*a[b];
        a[c]+=a[b];a[b]=0;
      }
      pl(sum+"");
    }
  }
  static final BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
  public static void pr(String str){
    System.out.print(str);
  }
  public static void pl(String str){
    System.out.println(str);
  }
  public static String read(){
    try{
      return ctos((char)br.read());
    }catch(IOException e){
      e.printStackTrace();
      return "";
    }
  }
  public static char readC(){
    try{
      return (char)br.read();
    }catch(IOException e){
      e.printStackTrace();
      return (char)-1;
    }
  }
  public static String readL(){
    try{
      return br.readLine();
    }catch(IOException e){
      e.printStackTrace();
      return "";
    }
  }
  public static String readS(){
    StringBuilder sb = new StringBuilder();
    while(true){
      try{
        int k = br.read();
        if(k==-1||(char)k==' '||(char)k=='\n'){break;}
        sb.append((char)k);
      }catch(IOException e){
        e.printStackTrace();
      }
    }
    return sb.toString();
  }
  public static int readI(){
    return stoi(readS());
  }
  public static long readLong(){
    return stol(readS());
  }
  public static long stol(String s){
    return Long.parseLong(s);
  }
  public static String[] readSs(){
    return readL().split(" ");
  }
  public static int[] readIs(){
    return stoi(readSs());
  }
  public static int stoi(String s){
    return Integer.parseInt(s);
  }
  public static int[] stoi(String s[]){
    int a[]=new int[s.length];
    for(int i=0;i<s.length;i++){
      a[i]=stoi(s[i]);
    }
    return a;
  }
  public static String itos(int i){
    return String.valueOf(i);
  }
  public static String[] itos(int[] a){
    String s[]=new String[a.length];
    for(int i=0;i<a.length;i++){
      s[i]=itos(a[i]);
    }
    return s;
  }
  public static String ctos(char c){
    return String.valueOf(c);
  }
  public static String cstos(char[] c){
    return new String(c);
  }
  public static char stoc(String s){
    return s.charAt(0);
  }
  public static char[] stocs(String s){
    return s.toCharArray();
  }
}