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 <iostream>
#include <iomanip>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <array>
#include <vector>
#include <utility>
#include <bitset>
#include <queue>
#include <unordered_map>
#include <unordered_set>
#include <numeric>

using namespace std;

using ll = long long;
using p = pair<ll, ll>;

constexpr ll INF = 1LL << 62;
constexpr ll MOD = 1e9 + 7;

int main(void) {
    cin.tie(0);
    ios::sync_with_stdio(false);

    ll H, W;
    cin >> H >> W;

    vector<ll> C(26, 0);
    for (ll h = 0; h < H; h++) {
        for (ll w = 0; w < W; w++) {
            char c;
            cin >> c;
            C[c - 'a']++;
        }
    }

    ll g2 = H / 2 + W / 2;
    ll g4 = (H / 2) * (W / 2);

    for (ll i = 0; i < 26; i++) {
        ll mod = C[i] % 4;
        if (mod == 1 || mod == 3) {
            C[i]--;
            break;
        }
    }

    for (ll k = 0; k < g2; k++) {
        for (ll i = 0; i < 26; i++) {
            ll mod = C[i] % 4;
            if (mod == 2) {
                C[i] -= 2;
                break;
            }
        }
    }

    for (ll k = 0; k < g4; k++) {
        for (ll i = 0; i < 26; i++) {
            ll mod = C[i] % 4;
            if (mod == 0) {
                C[i] -= 4;
                break;
            }
        }
    }

    for (ll i = 0; i < 26; i++) {
        if (C[i] != 0) {
            cout << "No" << endl;
            return 0;
        }
    }

    cout << "Yes" << endl;
    return 0;
}