Problem:
Score : 100 points

Problem Statement

Iroha loves Haiku. Haiku is a short form of Japanese poetry. A Haiku consists of three phrases with 5, 7 and 5 syllables, in this order.

To create a Haiku, Iroha has come up with three different phrases. These phrases have A, B and C syllables, respectively. Determine whether she can construct a Haiku by using each of the phrases once, in some order.

Constraints

1≦A,B,C≦10

Input

The input is given from Standard Input in the following format:

A B C

Output

If it is possible to construct a Haiku by using each of the phrases once, print YES (case-sensitive). Otherwise, print NO.

Sample Input 1

5 5 7

Sample Output 1

YES

Using three phrases of length 5, 5 and 7, it is possible to construct a Haiku.

Sample Input 2

7 7 5

Sample Output 2

NO

Solution in Rust (1.15.1) (judge verdict: Wrong Answer)

fn main() {
    let a: String = read();
    let a: usize = a.len();
    let b: String = read();
    let b: usize = b.len();
    let c: String = read();
    let c: usize = c.len();
    if max(a, max(b, c)) == 7 && min(a, min(b, c)) == 5 && a+b+c == 17 {
        println!("YES");
    } else {
        println!("NO");
    }
}

// 以下関数

use std::io::*;
use std::str::FromStr;
pub fn read<T: FromStr>() -> T {
    let stdin = stdin();
    let stdin = stdin.lock();
    let token: String = stdin
        .bytes()
        .map(|c| c.expect("failed to read char") as char)
        .skip_while(|c| c.is_whitespace())
        .take_while(|c| !c.is_whitespace())
        .collect();
    token.parse().ok().expect("failed to parse token")
}

pub const MOD: u64 = 1_000_000_007;

pub fn is_prime(n: i32) -> bool {
    if n < 2 {
        return false;
    } else if n == 2 {
        return true;
    }
    let mut i = 2;
    while i * i < n {
        if n % i == 0 {
            return false;
        } else {
            i += 1;
            continue;
        }
    }
    true
}

pub fn lcm(mut n: i32, mut m: i32) -> i32 {
    n = if n > m { m } else { n };
    m = if n > m { n } else { m };
    let mut i = n;
    while i % m != 0 {
        i += n;
    }
    i
}

pub fn abs(n: i32) -> i32 {
    if n >= 0 {
        n
    } else {
        -n
    }
}

pub fn max(n: usize, m: usize) -> usize {
    if n >= m {
        n
    } else {
        m
    }
}

pub fn min(n: usize, m: usize) -> usize {
    if n <= m {
        n
    } else {
        m
    }
}

pub fn factorial(n: u32) -> u32 {
    if n == 0 {
        return 1;
    }
    let mut r = n;
    for i in 0..n - 1 {
        r *= n - i - 1;
    }
    r
}

pub fn combination(n: u32, m: u32) -> u32 {
    factorial(n) / factorial(m) / factorial(n - m)
}