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;

fn main() {
    let mut is = String::new();
    stdin().read_line(&mut is).ok();
    let n: usize = is.trim().parse::<usize>().unwrap();
    is.clear();
    stdin().read_line(&mut is).ok();
    let mut a = is
        .split_whitespace()
        .map(|e| e.parse::<usize>().unwrap())
        .collect::<Vec<_>>();
    let mut dp = vec![0; 100001];

    for e in a {
        dp[e] += 1;
    }

    let mut ans = 0;

    for i in 1..dp.len() {
        ans += i * dp[i];
    }
    is.clear();
    stdin().read_line(&mut is).ok();
    let q: usize = is.trim().parse::<usize>().unwrap();
    for _ in 0..q {
        let mut is = String::new();
        stdin().read_line(&mut is).ok();
        let mut itr = is.split_whitespace().map(|e| e.parse::<usize>().unwrap());
        let b = itr.next().unwrap();
        let c = itr.next().unwrap();
        ans -= b * dp[b];
        ans -= c * dp[c];
        dp[c] += dp[b];
        dp[b] = 0;
        ans += c * dp[c];
        println!("{}", ans);
    }
}