Problem:
Score : 300 points

Problem Statement

Given is a sequence of integers A_1, A_2, ..., A_N.
If its elements are pairwise distinct, print YES; otherwise, print NO.

Constraints

2 ≤ N ≤ 200000

1 ≤ A_i ≤ 10^9

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
A_1 ... A_N

Output

If the elements of the sequence are pairwise distinct, print YES; otherwise, print NO.

Sample Input 1

5
2 6 1 4 5

Sample Output 1

YES

The elements are pairwise distinct.

Sample Input 2

6
4 1 3 1 6 2

Sample Output 2

NO

The second and fourth elements are identical.

Sample Input 3

2
10000000 10000000

Sample Output 3

NO

Solution in Rust (1.15.1) (judge verdict: Accepted)

macro_rules! input {
    (source = $s:expr, $($r:tt)*) => {
        let mut iter = $s.split_whitespace();
        input_inner!{iter, $($r)*}
    };
    ($($r:tt)*) => {
        let s = {
            use std::io::Read;
            let mut s = String::new();
            std::io::stdin().read_to_string(&mut s).unwrap();
            s
        };
        let mut iter = s.split_whitespace();
        input_inner!{iter, $($r)*}
    };
}

macro_rules! input_inner {
    ($iter:expr) => {};
    ($iter:expr, ) => {};
    ($iter:expr, $var:ident : $t:tt $($r:tt)*) => { let $var = read_value!($iter, $t); input_inner!{$iter $($r)*} }; }
macro_rules! read_value {
    ($iter:expr, ( $($t:tt),* )) => {
        ( $(read_value!($iter, $t)),* )
    };

    ($iter:expr, [ $t:tt ; $len:expr ]) => {
        (0..$len).map(|_| read_value!($iter, $t)).collect::<Vec<_>>()
    };

    ($iter:expr, chars) => {
        read_value!($iter, String).chars().collect::<Vec<char>>()
    };

    ($iter:expr, usize1) => {
        read_value!($iter, usize) - 1
    };

    ($iter:expr, $t:ty) => {
        $iter.next().unwrap().parse::<$t>().expect("Parse error")
    };
}

pub trait LexicalPermutation {
    /// Return `true` if the slice was permuted, `false` if it is already
    /// at the last ordered permutation.
    fn next_permutation(&mut self) -> bool;
    /// Return `true` if the slice was permuted, `false` if it is already
    /// at the first ordered permutation.
    fn prev_permutation(&mut self) -> bool;
}

impl<T> LexicalPermutation for [T]
where
    T: Ord,
{
    /// Original author in Rust: Thomas Backman <serenity@exscape.org>
    fn next_permutation(&mut self) -> bool {
        // These cases only have 1 permutation each, so we can't do anything.
        if self.len() < 2 {
            return false;
        }

        // Step 1: Identify the longest, rightmost weakly decreasing part of the vector
        let mut i = self.len() - 1;
        while i > 0 && self[i - 1] >= self[i] {
            i -= 1;
        }

        // If that is the entire vector, this is the last-ordered permutation.
        if i == 0 {
            return false;
        }

        // Step 2: Find the rightmost element larger than the pivot (i-1)
        let mut j = self.len() - 1;
        while j >= i && self[j] <= self[i - 1] {
            j -= 1;
        }

        // Step 3: Swap that element with the pivot
        self.swap(j, i - 1);

        // Step 4: Reverse the (previously) weakly decreasing part
        self[i..].reverse();

        true
    }

    fn prev_permutation(&mut self) -> bool {
        // These cases only have 1 permutation each, so we can't do anything.
        if self.len() < 2 {
            return false;
        }

        // Step 1: Identify the longest, rightmost weakly increasing part of the vector
        let mut i = self.len() - 1;
        while i > 0 && self[i - 1] <= self[i] {
            i -= 1;
        }

        // If that is the entire vector, this is the first-ordered permutation.
        if i == 0 {
            return false;
        }

        // Step 2: Reverse the weakly increasing part
        self[i..].reverse();

        // Step 3: Find the rightmost element equal to or bigger than the pivot (i-1)
        let mut j = self.len() - 1;
        while j >= i && self[j - 1] < self[i - 1] {
            j -= 1;
        }

        // Step 4: Swap that element with the pivot
        self.swap(i - 1, j);

        true
    }
}

#[warn(dead_code)]
fn sieve_of_eratosthenes(n: usize) -> Vec<usize> {
    // spf:
    // store smallest prime factor of number
    // 最小の素因数を格納
    // is_prime:
    // 配列番号が素数であるかどうかを示す
    let mut spf = vec![None; n + 1];
    let mut is_prime = vec![true; n + 1];
    let mut primes = Vec::new();

    // 0 と 1 は素数ではない
    is_prime[0] = false;
    is_prime[1] = false;

    for i in 2..n + 1 {
        // is_prime[i]:
        // それ以下の素数で割り切られていなければ True となる
        if is_prime[i] {
            // この if 文より i は素数なので
            primes.push(i);

            // 素数はそれ自身の最も小さい素因数であるから
            spf[i] = Some(i);
        }

        // 全ての prime の倍数(prime * i)を取り除く
        for prime in &primes {
            // prime が i の最も小さい因数よりも大きくなってしまったら探索終了
            if i * prime >= n + 1 || prime > &spf[i].unwrap() {
                break;
            }

            is_prime[i * prime] = false;

            // i * primes[j] の最も小さい素数は primes[j] 自身
            spf[i * prime] = Some(*prime);
        }
    }
    primes
}
use std::cmp::Ordering;
#[derive(Eq, PartialEq, Clone, Debug)]
pub struct Rev<T>(pub T);

impl<T: PartialOrd> PartialOrd for Rev<T> {
    fn partial_cmp(&self, other: &Rev<T>) -> Option<Ordering> {
        other.0.partial_cmp(&self.0)
    }
}

impl<T: Ord> Ord for Rev<T> {
    fn cmp(&self, other: &Rev<T>) -> Ordering {
        other.0.cmp(&self.0)
    }
}
fn gcd(a: u64, b: u64) -> u64 {
    if b == 0 {
        a
    } else {
        gcd(b, a % b)
    }
}
fn lcm(a: u64, b: u64) -> u64 {
    a / gcd(a, b) * b
}
use std::collections::BTreeSet;

#[warn(unused_mut)]
#[warn(unused_imports)]
fn main() {
    input! {
        n: u64,
        A: [u64;n],
    };

    let An: BTreeSet<_> = A.iter().collect();
    println!("{}", if n == An.len() as u64 { "YES" } else { "NO" });
}