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(unused, non_snake_case, dead_code, non_upper_case_globals)]
use {
    proconio::{marker::*, *},
    std::{cmp::*, collections::*, mem::*},
};
fn main() {
    input! {//
        h:usize,w:usize,
        hw:[Chars;h],
    }
    let mut v = vec![0usize; 26];
    for r in hw {
        for &c in &r {
            v[charconv(c, 'a')] += 1;
        }
    }
    let mut c = 0;
    if h % 2 == 1 {
        c += w / 2;
    }
    if w % 2 == 1 {
        c += h / 2;
    }
    let mut ret = 0;
    let mut ret2 = 0;
    for i in 0..26 {
        if v[i] % 4 == 2 {
            ret += 1;
        }
        if v[i] % 2 == 1 {
            ret2 += 1;
        }
    }
    if ret > c || ret2 > 1 {
        yn(false);
        return;
    }
    yn(true);
}
pub fn charconv(c: char, base: char) -> usize {
    return (c as u8 - base as u8) as usize;
}
pub fn yn(ans: bool) {
    if ans {
        println!("Yes");
    } else {
        println!("No");
    }
}