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)

#![allow(unused_imports)]

use std::cmp::{max, min, Ordering};
use std::collections::*;
use std::{i32, i64, i128, u32, u64, u128, f64};
use std::io::{stdout, Write, BufWriter};
use std::mem::{swap};
use proconio::marker::{Isize1, Usize1, Bytes, Chars};

#[proconio::fastout]
fn main() {
    proconio::input!{
        n: usize,
        arr: [i64; n],
        q: usize,
        bcs: [(i64, i64); q],
    }
    let mut counter = HashMap::new();
    for a in arr { *counter.entry(a).or_insert(0) += 1; }
    let mut state = counter.iter().map(|(k, v)| k * v).sum::<i64>();
    for (b, c) in bcs {
        if let Some(k) = counter.get_mut(&b) {
            state += (c - b) * *k;
            let b = *k;
            *k = 0;
            *counter.entry(c).or_insert(0) += b;
        }
        println!("{}", state);
    }
}