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 ll long long
#define mod 1000000007
#define endl "\n" 
#define p pair<int, int>
#define arrinp(i, a, start, end); for(ll i = start; i <= end; i++) cin >> a[i];
#define arrout(i, a, start, end); for(ll i = start; i <= end; i++) cout << a[i] << " ";
#define swap(a, b) {a = a + b; b = a - b; a = a - b;
#define forfront(i, start, end) for(ll i = start; i <= end; i++)
#define forback(i, start, end) for(ll i = start; i >= end; i--)
#define out(x) cout << x << " "
#define outline(x) cout << x << endl;
#define sortvector(v) sort(v.begin(), v.end());
#define sortvectorrev(v) sort(v.rbegin(), v.rend());
int main()
{
    #ifdef LOCAL
        freopen("input.txt", "r", stdin);
        freopen("output.txt", "w", stdout);
    #endif
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    ll n;
    cin >> n;
    ll a[n];
    arrinp(i, a, 0, n - 1);
    map<ll, ll> m;
    ll sum = 0;
    forfront(i, 0, n - 1)
    {
        m[a[i]]++;
        sum += a[i];
    }    
    ll q;
    cin >> q;
    while(q--)
    {
        ll b, c;
        cin >> b >> c;
        sum -= m[b]*b;
        sum += m[b]*c;
        ll count = m[b];
        m[b] = 0;
        m[c] += count;
        // out(c);
        // out(m[c]);
        outline(sum);
    }
    return 0;
}