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"

#define debug(x) cout<<#x<<": "<<x<<endl
#define rep(i,n) for (int i=0;i<(n);i++)
#define FOR(i,a,b) for (int i=(a);i<=(b);i++)
#define all(a) (a).begin(),(a).end()
using namespace std;
typedef vector<int> VI;
typedef vector<vector<int>> VVI;
typedef long long ll;


void solve() {
#ifdef _WIN32
	istream &cin = ifstream("input.txt");
#endif
	int h, w;
	cin >> h >> w;
	VI v(26);

	rep(i, h) {
		string s;
		cin >> s;
		rep(j, s.size()) {
			v[s[j] - 'a']++;
		}
	}
	sort(all(v));

	int num1 = ((w % 2 == 1 && h % 2 == 1) ? 1 : 0);
	int num2 = (w - 1) / 2 + (h - 1) / 2;
	int num4 = (w * h - num1 - num2 * 2) / 4;

	bool flag = false;
	for (int i = v.size() - 1; i >= 0; i--)
	{
		int tmp = min(num4, v[i] / 4);
		num4 -= tmp;
		v[i] -= tmp * 4;
		num2 -= v[i] / 2;
		v[i] -= v[i] / 2 * 2;
		num1 -= v[i];
		v[i] -= v[i];

		if (num4 == 0 && num2 == 0 && num1 == 0) {
			flag = true;
			break;
		}
	}

	

	cout << (flag ? "Yes" : "No") << endl;
}



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

	solve();
#ifdef _WIN32
	system("PAUSE");
#endif
	return 0;
}