Problem:
Score : 200 points

Problem Statement

Having learned the multiplication table, Takahashi can multiply two integers between 1 and 9 (inclusive) together.

Given an integer N, determine whether N can be represented as the product of two integers between 1 and 9. If it can, print Yes; if it cannot, print No.

Constraints

1 \leq N \leq 100

N is an integer.

Input

Input is given from Standard Input in the following format:

N

Output

If N can be represented as the product of two integers between 1 and 9 (inclusive), print Yes; if it cannot, print No.

Sample Input 1

10

Sample Output 1

Yes

10 can be represented as, for example, 2 \times 5.

Sample Input 2

50

Sample Output 2

No

50 cannot be represented as the product of two integers between 1 and 9.

Sample Input 3

81

Sample Output 3

Yes

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")};
}

fn main() {
    input! {n: u32}
    let v = (1..10).collect::<Vec<u32>>();
    for p in v.iter().rev() {
        if n % *p == 0 {
            if n / *p <= 9 {
                println!("Yes");
                return;
            }
            println!("No");
            return;
        }
    }
}