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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include "bits/stdc++.h"
#define in std::cin
#define out std::cout
#define rep(i,N) for(int i=0;i<N;++i)
typedef long long int LL;

int H, W;
std::string a[112];
int cou[100];
std::string field[112];

int main()
{
	in >> H >> W;
	rep(i, H)
	{
		in >> a[i];
		field[i].resize(W + 1);
	}

	rep(i, H)
	{
		for (auto c : a[i]) { ++cou[c - 'a']; }
	}
	rep(i, H)
	{
		rep(j, W)
		{
			if (field[i][j] == '\0')
			{
				int need = 0;

				/*if (W % 2 == 1 && H > 1 && (j + 1) == W / 2) { need = 2; }
				else if (H % 2 == 1 && W > 1 && (i + 1) == H / 2) { need = 2; }
				else if ((H % 2 == 1 || W % 2 == 1) && (i == H / 2 && j == W / 2)) { need = 1; }
				else if ((H == 1 && i == 0) || (W == 1 && j == 0)) { need = 2; }
				else { need = 4; }*/

				if (W == 1)
				{
					if (i == H / 2) { need = 1; }
					else { need = 2; }
				}
				else if (H == 1)
				{
					if (j == W / 2) { need = 1; }
					else { need = 2; }
				}
				else if (W % 2 == 1)
				{
					if (j == W / 2) { need = 2; }
					else { need = 4; }
				}
				else if (H % 2 == 1)
				{
					if (i == H / 2) { need = 2; }
					else { need = 4; }
				}
				else if (H % 2 == 1 && W % 2 == 1 && i == H / 2 && j == W / 2) { need = 1; }
				else { need = 4; }
				bool ok = false;
				for (char c = 'a'; c <= 'z'; ++c)
				{
					if (cou[c - 'a'] >= need)
					{
						ok = true;
						cou[c - 'a'] -= need;
						field[i][j] = c;
						field[H - 1 - i][j] = c;
						field[i][W - 1 - j] = c;
						field[H - 1 - i][W - 1 - j] = c;
						break;
					}
				}
				if (!ok)
				{
					out << "No" << std::endl;
					return 0;
				}
			}
		}
	}

	out << "Yes" << std::endl;
	return 0;
}