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)

//	Mohib Manva
#include<bits/stdc++.h>
using namespace std;
 
#define mod 1000000007
#define LOCAL 0
#define pb push_back
#define ll long long
 
ll po(ll a,ll b){
	ll x = 1,y=a;
	while(b>0){
		if(b%2){
			x = x*y;
			x %= mod;
		}
		y=y*y;
		y%=mod;
		b/=2;
	}
	return x;
}
 
int main(){
	if(LOCAL){
    	freopen("C:/Users/gold/Desktop/sublime IO/input.txt","r",stdin);
    	freopen("C:/Users/gold/Desktop/sublime IO/output.txt","w",stdout);
	}
	int T = 1;
	//scanf("%d",&T);
	while(T--){
		int n,m;
		scanf("%d %d",&n,&m);
		static char mat[105][105];
		for(int i=0;i<n;i++){
			scanf("%s",mat[i]);
		}		
		static int cnt[26];
		for(int i=0;i<n;i++){
			for(int j=0;j<m;j++){
				cnt[mat[i][j] - 'a']++;
			}
		}
		static int dp[4];
		for(int i=0;i<26;i++){
			if(cnt[i] % 4 == 0){
				continue;
			}
			dp[cnt[i]%4]++;
		}
		dp[1] += dp[3];
		dp[2] += dp[3];
        int u  = 0;
		if(dp[1] > 0){
			if(dp[1] == 1 && n % 2 && m % 2){
                u++;
			} else {
				puts("No");
				return 0;
			}
		}
		if(dp[2] > 0){
			int atmax = 0;
			if(n%2){
				atmax += m;
			}
			if(m%2){
				atmax += n;
			}
			atmax -= u;
			if(dp[2] <= atmax){
			} else {
				puts("No");
				return 0;
			}
		}
		puts("Yes");
	}
	return 0;	
}