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 ll long long
#define sz(x) (int)(x).size()
using namespace std;
//mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
//uniform_int_distribution<int>(1000,10000)(rng)

ll binpow(ll a, ll b) {
    if (b == 0)
    {
        return 1;
    }
    ll res = binpow(a, b / 2);
    res*=res;
    if (b % 2)
    {
        return res * a;
    }
    return res;
}

ll gcd(ll a,ll b)
{
    if (b==0) return a;
    return gcd(b,a%b);
}

string to_upper(string a)
{
    for (int i=0;i<(int)a.size();++i) if (a[i]>='a' && a[i]<='z') a[i]-='a'-'A';
    return a;
}
 
string to_lower(string a)
{
    for (int i=0;i<(int)a.size();++i) if (a[i]>='A' && a[i]<='Z') a[i]+='a'-'A';
    return a;
}

ll a[100001];

int main()
{
    ios_base::sync_with_stdio(0); cin.tie(0);
    int n,q,x,y;
    cin>>n;
    while (n--)
    {
        cin>>x;
        ++a[x];
    }
    ll ans=0;
    for (int i=1;i<=100000;++i)
        ans+=a[i]*i;
    cin>>q;
    while (q--)
    {
        cin>>x>>y;
        ans+=a[x]*(y-x);
        a[y]+=a[x];
        a[x]=0;
        cout<<ans<<"\n";
    }
    return 0;
}