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 (Clang 3.8.0) (judge verdict: Wrong Answer)

#include <algorithm>
#include <bitset>
#include <cassert>
#include <cmath>
#include <complex>
#include <cstdint>
#include <cstdio>
#include <deque>
#include <functional>
#include <iomanip>
#include <iostream>
#include <limits>
#include <map>
#include <numeric>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <tuple>
#include <utility>
#include <vector>
using namespace std;

using i64 = int64_t;

int H, W;
string s[101];

int main() {
    cin >> H >> W;
    for (int j = 0; j < H; ++j) {
        cin >> s[j];
    }

    int count[26];
    fill(count, count+26, 0);
    for (int j = 0; j < H; ++j) {
        for (int k = 0; k < W; ++k) {
            ++count[s[j][k]-'a'];
        }
    }

    if (H % 2 == 0 && W == 0) {
        for (int j = 0; j < 26; ++j) {
            if (count[j] % 4 == 0) {
                cout << "No" << endl;
                return 0;
            }
        }
        cout << "Yes" << endl;
        return 0;
    }

    if ((H % 2 == 0) != (W % 2 == 0)) {
        int w = (H % 2 == 1) ? W : H;
        int a = 0;
        for (int j = 0; j < 26; ++j) {
            if (count[j] % 2 == 1) {
                cout << "No" << endl;
                return 0;
            }
            if (count[j] % 4 == 2) {
                ++a;
            }
        }
        cout << (a <= w/2 ? "Yes" : "No") << endl;
        return 0;
    }

    int o = 0, d = 0;
    for (int j = 0; j < 26; ++j) {
        if (count[j] % 2 == 1) {
            ++o;
            count[j] %= 2;
        }
        if (count[j] % 4 == 2) {
            ++d;
        }
    }
    cout << ((o == 1 && d <= (H-1)/2+(W-1)/2) ? "Yes" : "No") << endl;
    return 0;
}