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)

#include<iostream>
#include<algorithm>
#include<vector>
#include<queue>
#define lol(i,n) for(int i=0;i<n;i++)
#define mod 1000000007
typedef long long ll;
using namespace std;
int cnt[26];
int main(){
	int h,w;
	lol(i,26)cnt[i]=0;
	cin>>h>>w;
	lol(i,h){
		lol(j,w){
			char z;
			cin>>z;
			cnt[z-'a']++;
		}
	}
	int d4=(h/2)*(w/2);
	int d2=0;
	if(h%2==1)d2+=w/2;
	if(w%2==1)d2+=h/2;
	int d1=0;
	if(h%2==1&&w%2==1)d1=1;
	bool ok=true;
	lol(i,26){
		while(d4){
			if(cnt[i]>=4){
				cnt[i]-=4;
			}
			else break;
			d4--;
		}
		while(d2){
			if(cnt[i]>=2){
				cnt[i]-=2;
			}
			else break;
			d2--;
		}
		if(cnt[i]==1){
			if(d1==1){
				cnt[i]=0;
				d1--;
			}
		}
		if(cnt[i]>0)ok=false;
	}
	if(ok)cout<<"Yes"<<endl;
	else cout<<"No"<<endl;
	return 0;
}