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)

#pragma GCC optimize ("O2")
#pragma GCC target ("avx2")
//#include<bits/stdc++.h>
#include<cstdio>
#include<cstring>

using namespace std;
typedef long long ll;
#define rep(i, n) for(int i = 0; i < (n); i++)
#define rep1(i, n) for(int i = 1; i <= (n); i++)
#define co(x) cout << (x) << "\n"
#define cosp(x) cout << (x) << " "
#define ce(x) cerr << (x) << "\n"
#define cesp(x) cerr << (x) << " "
#define pb push_back
#define mp make_pair
#define chmin(x, y) x = min(x, y)
#define chmax(x, y) x = max(x, y)
#define Would
#define you
#define please

const int CM = 1 << 17, CL = 12;
char cn[CM + CL], * ci = cn + CM + CL, * owa = cn + CM, ct;
const ll ma0 = 1157442765409226768;
const ll ma1 = 1085102592571150095;
const ll ma2 = 71777214294589695;
const ll ma3 = 281470681808895;
const ll ma4 = 4294967295;
inline int getint() {
	if (ci - owa > 0) {
		memcpy(cn, owa, CL);
		ci -= CM;
		fread(cn + CL, 1, CM, stdin);
	}
	ll tmp = *(ll*)ci;
	int dig = 68 - __builtin_ctzll((tmp & ma0) ^ ma0);
	tmp = tmp << dig & ma1;
	tmp = tmp * 10 + (tmp >> 8) & ma2;
	tmp = tmp * 100 + (tmp >> 16) & ma3;
	tmp = tmp * 10000 + (tmp >> 32) & ma4;
	ci += 72 - dig >> 3;
	return tmp;
}

//がんばってputintつくるかぁ
const int MAX = 10000;
const int dm = 1 << 17;
char dn[dm], * di = dn, * owad = dn + dm - 20;
class shuturyoku_unko {
public:
	char C[MAX * 4];
	shuturyoku_unko() : C() {
		rep(i, MAX) {
			int X = i;
			rep(j, 4) {
				C[i * 4 + 3 - j] = '0' + X % 10;
				X /= 10;
			}
		}
	}
	void putint(ll A) {
		if (owad < di) {
			fwrite(dn, 1, di - dn, stdout);
			di = dn;
		}
		int dig = 1;
		if (A >= 100000000) {
			if (A >= 10000000000) dig = 3;
			else if (A >= 1000000000) dig = 2;
			memcpy(di + 4 + dig, C + A % 10000 * 4, 4);
			A /= 10000;
			memcpy(di + dig, C + A % 10000 * 4, 4);
			A /= 10000;
			memcpy(di, C + A * 4 + 4 - dig, dig);
			di += 8 + dig;
		}
		else if (A >= 10000) {
			if (A >= 10000000) dig = 4;
			else if (A >= 1000000) dig = 3;
			else if (A >= 100000) dig = 2;
			memcpy(di + dig, C + A % 10000 * 4, 4);
			A /= 10000;
			memcpy(di, C + A * 4 + 4 - dig, dig);
			di += 4 + dig;
		}
		else {
			if (A >= 1000) dig = 4;
			else if (A >= 100) dig = 3;
			else if (A >= 10) dig = 2;
			memcpy(di, C + A * 4 + 4 - dig, dig);
			di += dig;
		}
		*di++ = '\n';
	}
} f;



int A[100001];
int main() {
	//cin.tie(0);
	//ios::sync_with_stdio(false);


	int N = getint();
	ll kotae = 0;
	rep(i, N) {
		int kari = getint();
		A[kari]++;
		kotae += kari;
	}

	int Q = getint();
	rep(i, Q) {
		int b = getint();
		int c = getint();

		A[c] += A[b];
		kotae += A[b] * ll(c - b);
		A[b] = 0;

		f.putint(kotae);
	}


	fwrite(dn, 1, di - dn, stdout);

	Would you please return 0;
}