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;

	typedef long long ll;

	typedef pair<int, int> pii;
	typedef pair<ll,ll> pll;

	typedef vector<int> vi;
	typedef vector<ll> vll;
	typedef vector<pii> vpii;
	typedef vector<pll> vpll;

	#define FOR(i, a, b) for (int i=a; i<(b); i++)
	#define F0R(i, a) for (int i=0; i<(a); i++)
	#define FORd(i,a,b) for (int i = (b)-1; i >= a; i--)
	#define F0Rd(i,a) for (int i = (a)-1; i >= 0; i--)

	#define ALL(s) (s).begin(),(s).end()
	#define ALLn(s,n) s,s+n
	#define F first
	#define S second
	#define pb push_back
	#define tc(t) int t; cin >> t; while(t--)
	#define _ ios_base::sync_with_stdio(false); cin.tie(NULL);
	#define D1(x) { cerr << " [" << #x << ": " << x << "]\n"; }
	#define D2(x) { cerr << " [" << #x << ": "; for(auto it:x) cerr << it << " "; cerr << "]\n"; }

	const ll MOD = 1e9 + 7;
	const ll MAXN = 1e6 + 7;
	const ll INF = LONG_LONG_MAX;

	void solve() {
		int n;
			cin >> n;
			int a[n];
			map<int,int> mp;
			ll sum = 0;

			F0R(i,n) {
				cin >> a[i];
				sum += a[i];
				mp[a[i]]++;
			}

			int q;
			cin >> q;

			F0R(i,q) {
				// for(auto x: mp) cout << x.F << ' ' <<x.S<<'\n';
				int b,c;
				cin >> b>> c;
				int diff = c - b;
				if(mp.count(c)==0) mp[c] = 0;
				mp[c] += mp[b];
				sum += (mp[b]*diff);
				mp[b]=0;
				cout << sum << '\n';
			}

	}

	int32_t main() {_
		
	#ifndef ONLINE_JUDGE
		freopen("in", "r", stdin);
		freopen("out", "w", stdout);
	#endif

		// tc(t)
			solve();
		cerr << "[ Time : " << (float)clock() / CLOCKS_PER_SEC << " secs ]" << endl;
	}