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 to edge[i].v
#define mp make_pair
#define rint register int
#define debug(x) cerr<<#x<<"="<<x<<endl
#define fgx cerr<<"-------------"<<endl
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
int num[10],s[100]; char c[1000][1000];
int main()
{	int n,m; cin>>n>>m;
	for(rint i=1;i<=n;i++) scanf("%s",c[i]+1);
	for(rint i=1;i<=n;i++)
	for(rint j=1;j<=m;j++)
	if(n-i+1!=i&&m-j+1!=j) num[0]++;
	else if(n-i+1==i&&m-j+1==j) num[1]++;
	else num[2]++;
	for(rint i=1;i<=n;i++)
	for(rint j=1;j<=m;j++) ++s[c[i][j]-'a'];
	for(rint i=0;i<26;i++)
	if(s[i])
	{	num[s[i]%4]--;
		if(num[s[i]%4]<0) return !printf("No");
	}
	printf("Yes");
	return 0;
}