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>
using namespace std;
using ll = long long;
#define REP(i, n) for (int i = 0; i < (int)(n); i++)
#define ALL(x) (x).begin(), (x).end()

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

    int H, W;
    cin >> H >> W;
    vector<string> a(H);
    REP(i, H) cin >> a[i];
    vector<int> cnt(26);
    REP(i, H) {
        REP(j, W) {
            cnt[a[i][j] - 'a']++;
        }
    }
    int g1, g2, g4;
    if (H % 2 == 0 && W % 2 == 0) {
        g1 = g2 = 0;
        g4 = (H / 2) * (W / 2);
    } else if (H % 2 == 1 && W % 2 == 0) {
        g1 = 0;
        g2 = W / 2;
        g4 = (H / 2) * (W / 2);
    } else if (H % 2 == 0 && W % 2 == 1) {
        g1 = 0;
        g2 = H / 2;
        g4 = (H / 2) * (W / 2);
    } else {
        g1 = 1;
        g2 = H / 2 + W / 2;
        g4 = (H / 2) * (W / 2);
    }
    REP(i, 26) {
        if (g1 > 0 && cnt[i] % 2 == 1) {
            cnt[i]--;
            g1--;
        }
    }
    REP(i, 26) {
        if (g2 > 0 && cnt[i] % 4 == 2) {
            cnt[i] -= 2;
            g2--;
        }
    }
    REP(i, 26) {
        if (g4 > 0 && cnt[i] > 0 && cnt[i] % 4 == 0) {
            cnt[i] -= 4;
            g4--;
        }
    }
    REP(i, 26) {
        g2 -= cnt[i] / 2;
    }
    if (g1 == 0 && g2 == 0 && g4 == 0) cout << "Yes" << endl;
    else cout << "No" << endl;
}