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)

#define _CRT_SECURE_NO_WARNINGS
#include <iostream>
#include <vector>
#include <deque>
#include <string>
#include <list>
#include <algorithm>
#include <cmath>
#include <fstream>
#include <map>
#include <set>
#include <sstream>
#include <ctime>
using namespace std;
typedef signed long long int lli;
#define FOR(i, p, n) for(int i(p); i < n; i++)
#define ALL(a) (a.begin()),(a.end())
#define sqrt(x) sqrt(1.0*(x))
#define pow(x,n) pow(1.0*(x),n)
#define UN(x) (x).resize(unique(ALL((x))) - (x).begin());
#define mp make_pair
const lli INF = 10e8;
map< char, int > mt;
void chet(int k){
	for (auto it : mt){
		int t = it.second;
		while (k && t>= 4){
			k--; t -= 4;
		}
		mt[it.first] = t;
	}
}
void dv(int k){
	for (auto it : mt){
		int t = it.second;
		while (k && t>= 2){
			k--;
			t -= 2;
		}
		mt[it.first] = t;
	}
}
bool solve()
{

	int n, m;
	cin >> n >> m;
	for (int i(0); i < n; i++){
		for (int j(0); j < m; j++){
			char t; cin >> t;
			mt[t]++;
		}
	}
	if ((n & 1) == 0 && (m & 1) == 0){
		for (auto it : mt){
			if (it.second % 4 != 0){
				cout << "No";
				return 1;
			}
		}
		cout << "Yes";
		return 1;
	}
	if (( n & 1)  && ( m & 1) ){
		chet((n-1)*(m-1)/4);
		dv((n+m-2)/2);
		for (auto it : mt){
			if (it.second == 1) {
				cout << "Yes";
				return 1;
			}

		}
		cout << "No";
		return 1;
	}
	if (( n & 1) == 1 && (m & 1) == 0){
		chet( ((n - 1)*m)/4);
		dv((m)/2);
		for (auto it : mt){
			if (it.second != 0){
				cout << "No";
				return 1;
			}
		}
		cout << "Yes";
		return 1;
	}
	if ((n & 1) == 0 && (m & 1) == 1){
		chet((n*(m - 1)) / 4);
		dv(n / 2);
		for (auto it : mt){
			if (it.second != 0){
				cout << "No";
				return 1;
			}
		}
		cout << "Yes";
	}
	return 1;
}
int main()
{
	solve();
	//while (solve());
	return 0;
}