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)

use proconio::input;
use proconio::marker::Chars;

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

    let mut map = std::collections::BTreeMap::new();
    for r in 0..h {
        for c in 0..w {
            *map.entry(a[r][c]).or_insert(0) += 1;
        }
    }
    let mut q = std::collections::BinaryHeap::new();
    for (k, v) in map {
        q.push((v, k));
    }
    let h_e = h % 2 == 0;
    let w_e = w % 2 == 0;
    if h_e && w_e {
        // ExE
        while let Some((v, k)) = q.pop() {
            if v < 4 {
                println!("No");
                return;
            }
            if v - 4 > 0 {
                q.push((v - 4, k));
            }
        }
        println!("Yes");
    } else if (h_e && !w_e) || (!h_e && w_e) {
        // ExO or OxE
        for _ in 0..if h_e { (w - 1) * h } else { (h - 1) * w } / 4 {
            if let Some((v, k)) = q.pop() {
                if v < 4 {
                    println!("No");
                    return;
                }
                if v - 4 > 0 {
                    q.push((v - 4, k));
                }
            }
        }
        // 1xE or Ex1
        while let Some((v, k)) = q.pop() {
            if v < 2 {
                println!("No");
                return;
            }
            if v - 2 > 0 {
                q.push((v - 2, k));
            }
        }
        println!("Yes");
    } else {
        // OxO
        if h == 1 && w == 1 {
            println!("Yes");
        } else if (h == 1 && w != 1) || (h != 1 && w == 1) {
            for _ in 0..std::cmp::max(h, w) / 2 {
                if let Some((v, k)) = q.pop() {
                    if v < 2 {
                        println!("No");
                        return;
                    }
                    if v - 2 > 0 {
                        q.push((v - 2, k));
                    }
                }
            }
            println!("Yes");
        } else {
            for _ in 0..(h - 1) * (w - 1) / 4 {
                if let Some((v, k)) = q.pop() {
                    if v < 4 {
                        println!("No");
                        return;
                    }
                    if v - 4 > 0 {
                        q.push((v - 4, k));
                    }
                }
            }
            for _ in 0..h / 2 + w / 2 {
                if let Some((v, k)) = q.pop() {
                    if v < 2 {
                        println!("No");
                        return;
                    }
                    if v - 2 > 0 {
                        q.push((v - 2, k));
                    }
                }
            }
            println!("Yes");
        }
    }
}