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)

//@sakshjha
#include<bits/stdc++.h>
using namespace std;
#define ll long long int
#define ld long double
#define FAST ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);
#define pb push_back
#define pi 3.14159265358979
const ll mod = 1000000007 ;
const ll N=500005;
int main()
{
    FAST
    ll t,i,j,k;
    //cin>>t;
    t=1;
    while(t--)
    {
        ll n,q;
        cin>>n;
        ll a[n];
        map<ll,ll>mp;
        ll sum=0;
        for(i=0; i<n; i++)
        {
            cin>>a[i];
            mp[a[i]]++;
            sum+=a[i];
        }
        cin>>q;
        while(q--)
        {
            ll u,v;
            cin>>u>>v;
            if(mp.find(u)==mp.end())
            {
                cout<<sum<<"\n";
            }
            else
            {
                ll cnt=mp[u];
                ll val=v-u;
                sum+=(cnt*val);
                mp.erase(u);
                mp[v]+=cnt;
                //cout<<cnt<<"\n";
                cout<<sum<<"\n";
            }
        }

    }
    return 0;
}
/*ll pow (ll x, ll y, ll m)
{
    ll res=1;
    x=x%m;
    if(x==0)
        return 0;
    while(y>0)
    {
        if(y&1)
        {
            res*=x;
            res%=m;
        }
        y=y/2;
        x=(x*x)%m;
    }
    return res;
}*/