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<stdio.h>
#include<string.h>
#include<algorithm>
#include<map>
#include<string>
using namespace std;
char a[110][110];
map<char,int> p;
int main()
{
    int i,j,k,n,m,flag,s1,s2,x1,x2,x3;
    scanf("%d%d",&n,&m);
    for(i=0;i<n;i++)
    scanf("%s",a[i]);
    for(i=0;i<n;i++)
    for(j=0;j<m;j++)
    p[a[i][j]]++;
    if(n==1&&m==1)
    {
        printf("Yes\n");
        return 0;
    }
    if(n==1)
    {
        k=strlen(a[0]);
        for(i=0;i<k/2;i++)
        {
            flag=0;
            for(j='a';j<='z';j++)
            if(p[j]>=2)
            {
                p[j]=p[j]-2;
                flag=1;
                break;
            }
            if(flag==0)
            break;
        }
        if(flag)
        printf("Yes\n");
        else
        printf("No\n");
        return 0;
    }
    if(m==1)
    {
        k=n;
        for(i=0;i<k/2;i++)
        {
            flag=0;
            for(j='a';j<='z';j++)
            if(p[j]>=2)
            {
                p[j]=p[j]-2;
                flag=1;
                break;
            }
            if(flag==0)
                break;
        }
        if(flag)
        printf("Yes\n");
        else
        printf("No\n");
        return 0;
    }
    s1=(n+1)/2;
    s2=(m+1)/2;
    for(i=0;i<s1;i++)
    {
        for(j=0;j<s2;j++)
        {
            int t1=n-i;
            int t2=m-j;
            int y1=4;
            if(t1==s1)
            y1=y1/2;
            if(t2==s2)
            y1=y1/2;
            flag=0;
            for(k='a';k<='z';k++)
            if(p[k]>=y1)
            {
                p[k]=p[k]-y1;
                flag=1;
                break;
            }
            if(flag==0)
                break;
        }
        if(flag==0)
            break;
    }
    if(flag)
        printf("Yes\n");
    else
        printf("No\n");
    return 0;
}