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 Rust (1.42.0) (judge verdict: Accepted)

use std::io::stdin;

#[derive(Default)]
struct Scanner {
	buffer: Vec<String>
}

impl Scanner {
	fn next<T: std::str::FromStr> (&mut self) -> T {
		loop {
			if let Some(token) = self.buffer.pop() {
				return token.parse().ok().expect("Failed parse");
			}
			let mut input = String::new();
			stdin().read_line(&mut input).expect("Failed read");
			self.buffer = input.split_whitespace().rev().map(String::from).collect();
		}
	}
}

fn main() {
	let mut scan = Scanner::default();
	let n: usize = scan.next();
	let mut a: [usize; 100005] = [0; 100005];
	let mut ans: usize = 0;
	let mut x: usize;
	let mut y: usize;
	for _ in 0..n {
		x = scan.next();
		a[x] += 1;
		ans += x;
	}
	let q: u32 = scan.next();
	for _ in 0..q {
		x = scan.next();
		ans -= x * a[x];
		y = a[x];
		a[x] = 0;
		x = scan.next();
		ans += x * y;
		a[x] += y;
		println!("{}", ans);
	}
}