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 REP(i, n) for(int i = 0; i < n; i++)
#define all(x) (x).begin(), (x).end()
#define INF 1000000007
#define mod 1005
#define ll long long int
const double mx = 2e6 + 5 ;

void solve()
{
	int n;
	cin>>n;
	map<ll, ll> mp;
	ll sum = 0;
	ll x;
	REP(i,n)
	{
		cin>>x;
		mp[x]++;
		sum+=x;
	}
	ll q, b,c;
	cin>>q;
	REP(i,q)
	{
		cin>>b>>c;
		sum-=mp[b]*b;
		sum+=mp[b]*c;
		mp[c]+=mp[b];
		mp.erase(b);
		cout<<sum<<"\n";
	}
}
int main()
{
	ios_base::sync_with_stdio(false);
    cin.tie(NULL);
	int t = 1;
//	cin>>t;
	while(t--)
	{
		solve();
	}
	return 0;
}