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)

/******************************************************************************

                              Online C++ Compiler.
               Code, Compile, Run and Debug C++ program online.
Write your code in this editor and press "Run" button to compile and execute it.

*******************************************************************************/

#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
const ll mod=1000000007;
#define MAX 50 



int main()
{
    std::ios::sync_with_stdio(false);
    std::cin.tie(0);

   
    ll n;
    cin>>n;
    ll a[n];
    map<ll,ll>m;
    ll sum=0;
    for(ll i=0;i<n;i++)
    {
        cin>>a[i];
        m[a[i]]++;
        sum+=a[i];
    }
    ll q;
    cin>>q;
    while(q--)
    {
        ll c=0;
        ll x,y;
        cin>>x>>y;
        //cout<<m[x]<<" "<<m[y];
        sum-=(m[x]*x);

        sum+=(m[x]*y);
        m[y]+=m[x];
        m[x]=0;

        /*for(int i=0;i<n;i++)
        {
            if(a[i]==x)
            a[i]=y;
            c+=a[i];
            
        }*/
        cout<<sum<<endl;
        
    }
    

    return 0;
}