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<vector>
#include<string>
#include<cstring>
#include<algorithm>
#include<map>
#include<set>
#include<cmath>
#include<cassert>
#include<queue>

using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;
typedef vector<int> vi;
struct cww{cww(){ios::sync_with_stdio(false);cin.tie(0);}}star;

void no() {
	cout << "No" << endl;
	exit(0);
}

bool isKai(const string& s) {
	int N = s.size();
	for (int i = 0; i < N/2; i++) {
		if (s[i] != s[N-1-i]) return false;
	}
	return true;
}

int main() {
	int H, W;
	cin >> H >> W;
	vector<string> board(H);
	for (int i = 0; i < H; i++)
		cin >> board[i];
	vector<vector<int> > hoge(26);
	for (int i = 0; i < W; i++) {
		hoge[board[0][i] - 'a'].push_back(i);
	}
	vector<int> perm(W);
	int oddCnt = 0;
	for (int i = 0; i < 26; i++) {
		if (hoge[i].size()%2) oddCnt++;
	}
	if (oddCnt >= 2) no();
	if (oddCnt > 0 && W%2 == 0) no();
	int left = 0, right = W;
	for (int i = 0; i < 26; i++) {
		int itr = 0;
		if (hoge[i].size() % 2) {
			perm[hoge[i][itr++]] = W/2;
		}
		for (; itr < hoge[i].size(); itr += 2) {
			perm[hoge[i][itr]] = left++;
			perm[hoge[i][itr+1]] = --right;
		}
	}
	// Debug
	//for (int i = 0; i < W; i++) {
	//	cout << perm[i];
	//}
	//cout << endl;
	map<string, int> mp;
	for (int i = 0; i < H; i++) {
		string next;
		for (int j = 0; j < W; j++) {
			next += " ";
		}
		for (int j = 0; j < W; j++) {
			next[perm[j]] = board[i][j];
		}
		if (!isKai(next)) no();
		if (mp.find(next) != mp.end()) mp[next]++;
		else mp[next] = 1;
	}
	oddCnt = 0;
	for (auto p : mp) {
		if (p.second%2) oddCnt++;
	}
	if (oddCnt > 1) no();
	if (oddCnt > 0 && H%2==0) no();
	cout << "Yes" << endl;
	return 0;
}