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 (1.15.1) (judge verdict: Wrong Answer)

fn main() {
  let (h, w): (usize, usize) = read_tuple();
  let mut mat: Vec<char> = read_vec();
  
  mat.sort();
  let mut freq = vec![0; 26];
  for c in mat {
    freq[c as usize - 'a' as usize] += 1;
  }
  
  let ans = match (h, w) {
    (1, 1) => true,
    (h, w) if h==1 && w%2==0 || h%2==0 && w==1 => freq.iter().all(|n| n%2==0),
    (h, w) if h==1 && w%2==1 || h%2==1 && w==1 => freq.iter().filter(|&n| n%2==1).count() == 1,
    (h, w) if h%2==0 && w%2==0 =>
      freq.iter().all(|n| n%4==0),
    (h, w) if h%2==1 && w%2==1 => 
      freq.iter().filter(|&&n| n>=4).count() >= std::cmp::max(h,w) / 2 &&
      freq.iter().filter(|&n| n%2==1).count() == 1,
    _ =>
      freq.iter().filter(|&&n| n>=4).count() >= std::cmp::max(h,w) / 2 &&
      freq.iter().all(|&n| n%2==0),
  };
  
  println!("{}", if ans {"Yes"} else {"No"});
}

fn read_tuple<T1: std::str::FromStr, T2: std::str::FromStr>() -> (T1, T2) {
  let mut buf = String::new();
  std::io::stdin().read_line(&mut buf).ok();
  let mut it = buf.trim().split_whitespace();
  let x = it.next().unwrap().parse::<T1>().ok().unwrap();
  let y = it.next().unwrap().parse::<T2>().ok().unwrap();
  (x, y)
}

use std::io::Read;
fn read_vec() -> Vec<char> {
  let mut buf = String::new();
  std::io::stdin().read_to_string(&mut buf).ok();
  buf.trim().chars().filter(|&c| char::is_alphabetic(c)).collect()
}