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 C# (Mono 4.6.2.0) (judge verdict: Wrong Answer)

using System;
using System.Linq;
using System.Collections.Generic;
class Program
{
	static void Main()
	{
	string[] input = Console.ReadLine().Split(' ');//Splitで区切り文字を指定して複数個受け取る。
	int h = int.Parse(input[0]);
	int w = int.Parse(input[1]);

    List<char> characters = new List<char>();
    List<int> numbers = new List<int>();//文字の数一覧
    for(int i = 0; i < h; i++)
    {
    string numsA = Console.ReadLine();
    char[] nums = numsA.ToCharArray();
      for(int j = 0; j < nums.Length; j++) characters.Add(nums[j]);
    }
    characters.Sort();//並び替え

    int count = 1;//文字数を数える
    for(int i = 0; i < characters.Count()-1; i++)
    {
      if(characters[i] == characters[i+1])
      {
      count++;
        if(i == characters.Count()-2) numbers.Add(count);
      }
      else
      {
        numbers.Add(count);
        if(i == characters.Count()-2) numbers.Add(1);
        count = 1;
      }
    }
    numbers.Sort();

    

    int oddNum = 0;//奇数の個数
    for(int i = 0; i < numbers.Count(); i++)
    {
      if(numbers[i] % 2 == 1) oddNum++;
      numbers[i] = numbers[i]-1;
    }

    int secondNum = 0;//2(mod4)の個数
    for(int i = 0; i < numbers.Count(); i++)
    {
      if(numbers[i] % 4 == 2) secondNum++;
      numbers[i] = numbers[i]-2;
    }

    bool answer = true;

    if((h == 1 && w % 2 == 0)||(w == 1 && h % 2 == 0))
    {
        if(oddNum >= 1) answer = false;
    }
    
    else if((h == 1 && w % 2 == 1)||(w == 1 && h % 2 == 1))
    {
      if(oddNum >= 2) answer = false;
    }

    else if(h % 2 == 0 && w % 2 == 0)
    {
      if(oddNum >= 1 || secondNum >= 1) answer = false;
    }

    else if(h % 2 == 0 && w % 2 == 1)
    {
      if(oddNum >= 1 || secondNum > h) answer = false;
    }

    else if(h % 2 == 1 && w % 2 == 0)
    {
      if(oddNum >= 1 || secondNum > w) answer = false;
    }

    else if(h % 2 == 1 && w % 2 == 1)
    {
      if(oddNum >= 2 || secondNum > h/2+w/2) answer = false;
    }

    Console.WriteLine(answer ? "Yes" : "No");

	}
}