Problem:
Score : 400 points

Problem Statement

We have an H-by-W matrix.
Let a_{ij} be the element at the i-th row from the top and j-th column from the left.
In this matrix, each a_{ij} is a lowercase English letter.

Snuke is creating another H-by-W matrix, A', by freely rearranging the elements in A.
Here, he wants to satisfy the following condition:

Every row and column in A' can be read as a palindrome.

Determine whether he can create a matrix satisfying the condition.

Note

A palindrome is a string that reads the same forward and backward.
For example, a, aa, abba and abcba are all palindromes, while ab, abab and abcda are not.

Constraints

1 ≤ H, W ≤ 100

a_{ij} is a lowercase English letter.

Input

Input is given from Standard Input in the following format:

H W
a_{11}a_{12}...a_{1W}
:
a_{H1}a_{H2}...a_{HW}

Output

If Snuke can create a matrix satisfying the condition, print Yes; otherwise, print No.

Sample Input 1

3 4
aabb
aabb
aacc

Sample Output 1

Yes

For example, the following matrix satisfies the condition.

abba
acca
abba

Sample Input 2

2 2
aa
bb

Sample Output 2

No

It is not possible to create a matrix satisfying the condition, no matter how we rearrange the elements in A.

Sample Input 3

5 1
t
w
e
e
t

Sample Output 3

Yes

For example, the following matrix satisfies the condition.

t
e
w
e
t

Sample Input 4

2 5
abxba
abyba

Sample Output 4

No

Sample Input 5

1 1
z

Sample Output 5

Yes

Solution in Rust (1.42.0) (judge verdict: Accepted)

#![allow(clippy::needless_range_loop)]
#![allow(unused_macros)]
#![allow(dead_code)]
#![allow(unused_imports)]

use itertools::Itertools;
use proconio::input;
use proconio::marker::*;

fn main() {
    input! {
        h: usize,
        w: usize,
        a: [Bytes; h]
    }

    let pairs = (h % 2) * (w / 2) + (w % 2) * (h / 2);
    let single = (h % 2) * (w % 2);

    let mut char_count = [0; 26];
    for row in &a {
        for &byte in row {
            char_count[(byte - b'a') as usize] += 1;
        }
    }

    let odd_cnt = char_count.iter().filter(|&&x| x % 2 != 0).count();
    let half_odd_cnt = char_count.iter().filter(|&&x| x / 2 % 2 != 0).count();

    let ans = odd_cnt == single && half_odd_cnt <= pairs && half_odd_cnt % 2 == pairs % 2;

    if ans {
        println!("Yes");
    } else {
        println!("No");
    }
}