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: Accepted)

#include <bits/stdc++.h>
using namespace std;
typedef int64_t ll;
typedef vector<string> vss;
struct Solution {
    void solve(std::istream& in, std::ostream& out) {
        ll h, w;
        in >> h >> w;
        vss a(h);
        ll count[26] = {};
        for (int i = 0; i < h; i++) {
            in >> a[i];
            for (int j = 0; j < w; j++) {
                count[a[i][j] - 'a']++;
            }
        }
        ll g1 = 1;
        ll g2 = h / 2 + w / 2;
        ll g4 = (h / 2) * (w / 2);
        if (w % 2 == 0 && h % 2 == 0) {
            while (g4-- > 0) {
                for (int i = 0; i < 26; ++i) {
                    if (count[i] % 4 == 0 && count[i]) {
                        count[i] -= 4;
                        break;
                    }
                }
            }
        }
        else if (w % 2 == 0 || h % 2 == 0) {
            bool flag = false;
            if (h % 2 == 1) {
                g2 = h == 1 ? w : w / 2;
                flag = h == 1;
            }
            else if (w % 2 == 1) {
                g2 = w == 1 ? h : h / 2;
                flag = w == 1;
            }
            while (g2-- > 0) {
                for (int i = 0; i < 26; ++i) {
                    if ((flag && count[i] % 2 == 0 && count[i]) || count[i] % 4 == 2) {
                        count[i] -= 2;
                        break;
                    }
                }
            }
            while (g4-- > 0) {
                for (int i = 0; i < 26; ++i) {
                    if (count[i] % 4 == 0 && count[i]) {
                        count[i] -= 4;
                        break;
                    }
                }
            }
        }
        else {
            bool flag = false;
            if (h == 1) {
                g2 = w;
                flag = h == 1;
            }
            else if (w == 1) {
                g2 = h;
                flag = w == 1;
            }
            while (g1-- > 0) {
                for (int i = 0; i < 26; ++i) {
                    if (count[i] % 4 == 1 || count[i] % 4 == 3) {
                        count[i]--;
                        break;
                    }
                }
            }
            while (g2-- > 0) {
                for (int i = 0; i < 26; ++i) {
                    if ((flag && count[i] % 2 == 0 && count[i]) || count[i] % 4 == 2) {
                        count[i] -= 2;
                        break;
                    }
                }
            }
            while (g4-- > 0) {
                for (int i = 0; i < 26; ++i) {
                    if (count[i] % 4 == 0 && count[i]) {
                        count[i] -= 4;
                        break;
                    }
                }
            }
        }
        ll sum = 0;
        for (int i = 0; i < 26; ++i) {
            sum += count[i];
        }
        out << (sum == 0 ? "Yes" : "No") << '\n';
    }
};
void solve(std::istream& in, std::ostream& out) {
    out << std::setprecision(12);
    Solution solution;
    solution.solve(in, out);
}
#include <fstream>
#include <iostream>
int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    istream& in = cin;
    ostream& out = cout;
    solve(in, out);
    return 0;
}