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: Accepted)

#include <bits/stdc++.h>
#define rep2(i,m,n) for (int i = (int)(m); i < (int)(n); i++)
#define rep(i,n) rep2(i,0,n)
#define all(x) (x).begin(), (x).end()
#define sz(x) ((int)(x).size())
#define pb push_back
#define bg begin()
#define ed end()
#define fi first
#define se second
using namespace std;
using ll = long long;
using vint = vector<int>;
using pint = pair<int,int>;
using vpint = vector<pint>;
template <typename T>
T sq(T x){
  return x * x;
}
ll mod = 1e9 + 7;

signed main (){
 int N; 
cin >> N;
vint a(N);
vint table(101010);
ll sum_i =0;
rep(i,N){
  cin >>a.at(i);
  ++table.at(a.at(i));
sum_i += a.at(i);
}
int Q;
cin >> Q;
vpint P(Q);
rep(i,Q){
int B,C;
  cin >> B >> C;
  P.at(i) = make_pair(B,C);
}
rep(i,Q){
// cout << P.at(i).fi << P.at(i).se<< endl;
// cout << table.at(P.at(i).first) << endl;
ll ans;
  if(table.at(P.at(i).fi) > 0){
  ans = sum_i - (P.at(i).fi - P.at(i).se )*table.at(P.at(i).fi );
sum_i = ans;
int num = table.at(P.at(i).fi);
  table.at(P.at(i).fi)-= num;
table.at(P.at(i).se) += num;
}else{
ans = sum_i;
}

  cout << ans <<endl;
}


    return 0;
}