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>
using namespace std;
#define pi 2*asin(1.0)
 
int main() {

int N;
long Ans=0;
int Bs;
cin>>N;
vector<int> A(N);
vector<int> D(100001,0);
for(int i=0;i<N;i++){
cin>>A.at(i);
D.at(A.at(i))++;
Ans=Ans+A.at(i);
}
int Q;
cin>>Q;
vector<int> B(Q);
vector<int> C(Q);
vector<long> S(Q);
for(int i=0;i<Q;i++){
cin>>B.at(i);
cin>>C.at(i);
}
for(int i=0;i<Q;i++){
    Bs=D.at(B.at(i));
    Ans=Ans+Bs*(C.at(i)-B.at(i));
    cout<<Ans<<endl;
	D.at(C.at(i))=D.at(B.at(i))+D.at(C.at(i));
    D.at(B.at(i))=0;
}




}