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>
using namespace std;
typedef long long ll;
const int MOD = 1e9+7,MAX = 1e5;

int main()
{
	int h,w;
	cin >> h >> w;
	vector<string> a(h);
	for(int i = 0;i<h;i++)cin >> a[i];
	vector<int> cnt(26);
	for(int i=0;i<h;++i)
	{
		for(int j = 0;j<w;++j)
		{
			cnt[a[i][j]-'a']++;
		}
	}
	
	for(int i =0;i<26;i++)cnt[i]%=4;
	//両方奇数奇数がひとつだけでおｋ
	if((h%2==1)&&(w%2==1))
	{
		ll is  = 0;
		for(int i = 0;i<26;i++)is+=cnt[i]&1;
		if(is<=1)
		{
			cout<<"Yes"<<endl;
		}
		else
		{
			cout<<"No"<<endl;
		}
		
	}//片方偶数奇数が残ってたらアウト
	else if((h%2==1&&w%2==0)||(h%2==0&&w%2==1))
	{
		//cout<<';'<<endl;
		ll c = 0;
		for(int i = 0;i<26;i++)
		{
			c += cnt[i]/2;
			if(cnt[i]%2==1)
			{
				cout<<"No"<<endl;
				return 0;
			} 
		}
		ll gusu = h&1?w:h;
		//cout<<gusu<<endl;
		//cout<<c<<endl;
		if(c>gusu/2)
		{
			cout<<"No"<<endl;
		}
		else
		{
			cout<<"Yes"<<endl;
		}
		

	}
	else//両方偶数全部ゼロ
	{
		bool is = true;
		for(int i =0;i<26;i++)
		{
			if(cnt[i])
			{
				cout<<"No"<<endl;
				return 0;
			}
		}
		cout<<"Yes"<<endl;


	}
	return 0;
}