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: Accepted)

/*input
3 4
aabb
aabb
aacc
*/
#include <bits/stdc++.h>
using namespace std;
const long long md=1e9+7;
long long a[1000005],n,freq[150];

int main(){
	#ifdef ONLINE_JUDGE
	ios_base::sync_with_stdio(0);
	cin.tie(0);cout.tie(0);
	#endif
	long long q,m,k,h,w,ans=0,sum=0,p,x=0,y=0;
	cin>>h>>w;
	string s;
	for(int i=0;i<h;++i){
		cin>>s;
		for(int j=0;j<s.size();++j)
			++freq[s[j]];
	}
	if(h%2==0 && w%2==0){
		for(int i='a';i<='z';++i)
			if(freq[i]%4) return cout<<"No",0;
		cout<<"Yes";
	}
	else if(h%2 && w%2){
		for(int i='a';i<='z';++i)
			if(freq[i]%4==3) ++sum,++x;
			else if(freq[i]%4==1) ++sum;
			else if(freq[i]%4==2) ++x;
		cout<<(sum==1 && x<=(h+w-2)/2&&((h+w)/2-1-x)%2==0?"Yes":"No");
	}
	else if(h%2 && w%2==0){
		for(int i='a';i<='z';++i)
			if((freq[i]%4)%2) return cout<<"No",0;
			else if(freq[i]%4==2) ++x;
		cout<<(x<=w/2&&(w/2-x)%2==0?"Yes":"No");
	}
	else{
		for(int i='a';i<='z';++i)
			if((freq[i]%4)%2) return cout<<"No",0;
			else if(freq[i]%4==2) ++x;
		cout<<(x<=h/2&&(h/2-x)%2==0?"Yes":"No");
	}
}