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 <iostream>
using namespace std;

const int N=1e5+10;

int n,q,x,y;
int a[N];
int b[N];
long long sum;

int main(){
	scanf("%d", &n);
	
	for(int i=0;i<n;i++) scanf("%d",&a[i]), b[a[i]]++, sum+=a[i];
	
	scanf("%d", &q);
	while(q--){
		scanf("%d%d",&x,&y);
		sum+=(y-x)*b[x];
		b[y]+=b[x],b[x]=0;
		printf("%ld\n",sum);
	}
	return 0;
}