Problem: 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.

Examples

Input

4
1 2 3 4
3
1 2
3 4
2 4


Output

11
12
16


Input

4
1 1 1 1
3
1 2
2 1
3 5


Output

8
4
4


Input

2
1 2
3
1 100
2 100
100 1000


Output

102
200
2000

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
#define int long long
using namespace std;
int times[100001],n,q,sum;
signed main()
{
	scanf("%lld",&n);
	for(int i=1;i<=n;i++)
	{
		int x;
		scanf("%lld",&x);
		times[x]++;
		sum+=x;
	}
	scanf("%lld",&q);
	while(q--)
	{
		int x,y;
		scanf("%lld%lld",&x,&y);
		sum+=(y-x)*times[x];
		cout<<sum<<endl;
		times[y]+=times[x];
		times[x]=0;
	}
	return 0;
}