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 fastinput   ios_base::sync_with_stdio(false),cin.tie(NULL),cout.tie(NULL)
#define ll          long long
#define bn          1000090
#define sn          1010
#define mod         100000007
#define iii(i)      cin>>i
#define ooo(i)      cout<<i
#define sss         cout<<" "
#define nnn         cout<<"\n"/**
#define iii(i)      iii>>i
#define ooo(i)      ooo<<i
#define sss         ooo<<" "
#define nnn         ooo<<"\n"*/
#define ff          first
#define ss          second
#define psb         push_back

/**
        ifstream iii;iii.open("D:\\prgrms\\codechef cntst 1\\input.txt");
        ofstream ooo;ooo.open("D:\\prgrms\\codechef cntst 1\\my output.txt");

		ooo("a a["<<n<<"] = "<<a[n]<<"  yom["<<y<<"] = "<<m[y]);nnn;
*/


ll a[bn];
map<ll,ll>m;

int main()
{
	///fastinput;
	ll n; iii(n);
	for(ll i = 1; i <= n; i++)
	{
		iii(a[i]);
		m[a[i]]++;
		a[i] += a[i-1];
	}
	ll q; iii(q);
	for(ll i = 1; i <= q; i++)
	{
		ll x,y;
		iii(x>>y);
		a[n] += m[x]*(y-x);
		ooo(a[n]); nnn;
		m[y] += m[x]; m[x] = 0;
	}
        return 0;
}




/**



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












*/















/***/