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 rep(i,n) for(int i=0; i<(n); i++)
#define rep2(i,x,n) for(int i=x; i<(n); i++)
#define all(n) begin(n),end(n)
struct cww{cww(){ios::sync_with_stdio(false);cin.tie(0);}}star;
const long long INF = numeric_limits<long long>::max();
int main(){
    int N;
    cin>>N;
    vector<int> A;
    long long ans=0;
    rep(i,N){
       int tmp;
       cin>>tmp;
       A.push_back(tmp); 
       ans+=tmp;
    }
    int Q;
    cin>>Q;
    vector<int> B,C;
    rep(i,Q){
        int b,c;
        cin>>b>>c;
        B.push_back(b);
        C.push_back(c);
    }
    map<int,int> nums;//(数、個数)で持つ
    rep(i,N){
        if(nums.count(A[i])){
            nums[A[i]]++;
        }else{
            nums[A[i]]=1;
        }
    }

    rep(i,Q){
        //各クエリでmapを更新、何が0になって何が増えたのか、総和はどうなったかを更新
        if(!nums.count(B[i])){
            //Biに相当する数が数列上にいないなら前のans引継ぎ
            cout << ans << endl;
            continue;
        }
        if(B[i]>C[i]){
            //総和は減るはず
            ans=ans-(nums[B[i]]*B[i])+nums[B[i]]*C[i];//Biがnums[Bi]個へってCiがnums[B[i]]こ増えた
            cout << ans << endl;
                if(nums.count(C[i])){
                nums[C[i]]+=nums[B[i]];//いるならBiの個数を足す
                }else{
                    nums[C[i]]=nums[B[i]];//いないならBiそのもの
                }
                //Biの個数は0に
                nums[B[i]]=0;

        }else{
        //総和はふえるはず
            ans=ans-(nums[B[i]]*B[i])+nums[B[i]]*C[i];//Biがnums[Bi]個へってCiがnums[B[i]]こ増えた
            cout << ans << endl;
                if(nums.count(C[i])){
                nums[C[i]]+=nums[B[i]];//いるならBiの個数を足す
                }else{
                    nums[C[i]]=nums[B[i]];//いないならBiそのもの
                }
                //Biの個数は0に
                nums[B[i]]=0;   
        }

    }



    return 0;
}