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<bits/stdc++.h>
using namespace std;
inline long long getnum()
{
    long long now=0;long long fh=1;char ch=getchar();
    while(ch<'0' || ch>'9'){if(ch=='-')fh=-1;ch=getchar();}
    while(ch>='0' && ch<='9'){now=(now<<3)+(now<<1)+ch-'0';ch=getchar();}
    return now*fh;
}
int ch[128];
int n,m;
int md[5];
int main()
{
  n=getnum();
  m=getnum();
  for (int i=1;i<=n*m;i++)
  {
    char tmp=getchar();
    while(tmp<'a'||tmp>'z')
    {
      tmp=getchar();
    }
    ch[(int)tmp]++;
  }
  for (int i='a';i<='z';i++)
  {
    md[ch[i]%4]++;
  }
  if (md[3]!=0)
  {
    printf("No\n");
    return 0;
  }
  if (n%2==0&&m%2==0)
  {
    if (md[1]>0||md[2]>0) printf("No\n"); else printf("Yes\n");
    return 0;
  }
  if (n%2!=0&&m%2!=0)
  {
    if (md[1]>1||md[2]>(n/2+m/2)) printf("No\n"); else printf("Yes\n");
    return 0;
  }
  if (n%2==0)
  {
    if (md[1]>0||md[2]>(n/2))printf("No\n");else printf("Yes\n");
    return 0;
  }else
  {
    if (md[1]>0||md[2]>(m/2))printf("No\n");else printf("Yes\n");
  }
  return 0;
}