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 rep(i, n) for (lli i = 0; i < (n); i++)
#define rrep(i, n) for (lli i = (n)-1; i >= 0; i--)
using namespace std;
using lli = long long int;
int main()
{
    string s;
    int n, m;
    cin >> n >> m;
    int cnt[30] = {};
    if (m & 1)
        swap(n, m);
    rep(i, n) rep(j, m)
    {
        char a;
        cin >> a;
        cnt[a - 'a']++;
    }
    rep(i, 26)
    {
        cnt[i] %= 4;
    }


    int c = 0;
    if (n & 1 and m & 1) {
        rep(i, 26)
        {
            if (cnt[i] == 1 or cnt[i] == 3) {
                c++;
            }
        }
        if (c > 1) {
            cout << "No" << endl;
        } else {
            cout << "Yes" << endl;
        }
    } else if (n & 1) {
        rep(i, 26)
        {
            if (cnt[i] == 1 or cnt[i] == 3) {
                c++;
            }
        }
        if (c) {
            cout << "No" << endl;
        } else {
            cout << "Yes" << endl;
        }
    } else {
        rep(i, 26)
        {
            if (cnt[i] == 1 or cnt[i] == 3 or cnt[i] == 2) {
                c++;
            }
        }
        if (c) {
            cout << "No" << endl;
        } else {
            cout << "Yes" << endl;
        }
    }
}