Problem:
Score : 500 points

Problem Statement

You are given a sequence with N integers: A = \{ A_1, A_2, \cdots, A_N \}.
For each of these N integers, we will choose a color and paint the integer with that color. Here the following condition must be satisfied:

If A_i and A_j (i < j) are painted with the same color, A_i < A_j.

Find the minimum number of colors required to satisfy the condition.

Constraints

1 \leq N \leq 10^5

0 \leq A_i \leq 10^9

Input

Input is given from Standard Input in the following format:

N
A_1
:
A_N

Output

Print the minimum number of colors required to satisfy the condition.

Sample Input 1

5
2
1
4
5
3

Sample Output 1

2

We can satisfy the condition with two colors by, for example, painting 2 and 3 red and painting 1, 4, and 5 blue.

Sample Input 2

4
0
0
0
0

Sample Output 2

4

We have to paint all the integers with distinct colors.

Solution in Rust (1.42.0) (judge verdict: Wrong Answer)

use std::cmp;
use std::collections::BTreeSet;
use std::collections::VecDeque;
use std::fmt::Debug;
use std::str::FromStr;
#[allow(dead_code)]
fn read_as_vec<T>() -> Vec<T>
where
    T: FromStr,
    <T as FromStr>::Err: Debug,
{
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).unwrap();
    s.trim()
        .split_whitespace()
        .map(|c| T::from_str(c).unwrap())
        .collect()
}

#[allow(dead_code)]
fn read_as_string() -> String {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).unwrap();
    s.trim().to_string()
}

fn lower_bound<T: Ord>(a: &VecDeque<T>, t: T) -> usize {
    let (mut left, mut right) = (0, a.len());
    while left < right {
        let mid = (left + right) / 2;
        if t <= a[mid] {
            left = mid + 1;
        } else {
            right = mid;
        }
    }
    return right;
}

fn main() {
    let n: usize = read_as_vec()[0];
    let mut a: VecDeque<usize> = VecDeque::new();

    for _ in 0..n {
        let t = read_as_vec::<usize>()[0];
        let pos = lower_bound(&a, t);
        if pos == a.len() {
            a.push_front(t);
        } else {
            a[pos] = t;
        }
    }
    println!("{}", a.len());
}