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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define ld long double
const int MOD=1e9+7;
const long double Pi=acos(-1);
using  P=pair<ll,ll>;
ll gcd(ll a,ll b){//aとbの最大公約数を返す
  if(a%b==0){return b;}
  else{return gcd(b,a%b);}
}
ll lcm(ll a,ll b){//aとbの最小公倍数を返す
  return a*b/gcd(a,b);
}
ll pow(ll x,ll y){//x^yを返す
  ll tmp=x,res=1;
  while(y){
    if(y%2){res=res*tmp%MOD;}
    y>>=1;
    tmp=tmp*tmp%MOD;
  }
  return res;
}
ll nature(ll a){//絶対値を返す
  if(a>=0){return(a);}
  else{return(-1*a);}
}
ll prime(ll a){//素数なら1を返す
  if(a==2){return 1;}
  else if(a%2==0 || a==1){return 0;}
  for(ll i=3;i*i<=a;i+=2){if(a%i==0){return 0;}}
  return 1;
}
int main(){
  int N,sum=0;
  cin>>N;
  vector<int> A(N);
  unordered_map<int,int> mp;
  for(int i=0;i<N;i++){
    cin>>A.at(i);
    mp[A.at(i)]++;
    sum+=A.at(i);
  }
  int Q;
  cin>>Q;
  for(int i=0;i<Q;i++){
    int a,b;
    cin>>a>>b;
    mp[b]+=mp[a];
    sum-=a*mp[a];
    sum+=b*mp[a];
    mp[a]=0;
    cout<<sum<<endl;
  }
}