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 long long ll;
typedef vector<ll> vl;
typedef vector<string> vs;
typedef pair<ll, ll> pair_t;
using Graph = vector<vector<ll>>;

#define FOR(i, a, b) for (ll i = (a); i < (b); i++)
#define REP(i, n) FOR(i, 0, n)
/***** define constant start *******/
const ll MOD = 1000000007;
const double PI = 3.1415926535897932;
const ll INF = 1 << 29;
/****** define constant end ********/
/****** define variable start ******/
ll h, w;
/******* define variable end *******/
ll solve() {
	cin >> h >> w;
	vector<vector<char>> a(h, vector<char>(w));
	vector<ll> count(26);
	REP(i, h){
		REP(j, w){
			cin >> a[i][j];
			int index = a[i][j] - 'a';
			count[index]++;
		}
	}
	vector<ll> count_vec;
	REP(i, 26){
		if(count[i] != 0){
			count_vec.push_back(count[i]);
		}
	}
	ll four_count = 0;
	ll two_count = 0;
	ll one_count = 0;
	REP(i, count_vec.size()){
		if(count_vec[i] >= 4){
			four_count++;
			count_vec[i] -= 4;
			i--;
		}
		else if(count_vec[i] >= 2){
			two_count++;
			count_vec[i] -= 2;
			i--;
		}
		else if(count_vec[i] >= 1){
			one_count++;
			count_vec[i] -= 1;
			i--;
		}
	}
	if(h % 2 == 0 && w % 2 == 0){
		if(four_count >= h * w / 4){
			cout << "Yes" << endl;
		}else{
			cout << "No" << endl;
		}
	}else if(h % 2 == 0 && w % 2 == 1){
		if(four_count >= h/2 * (w-1)/2 && two_count + (four_count - h/2 * (w-1)/2) * 2== h/2){
			cout << "Yes" << endl;
		}else{
			cout << "No" << endl;
		}
	}else if(h % 2 == 1 && w % 2 == 0){
		if(four_count >= (h-1)/2 * w/2 && two_count  + (four_count - (h-1)/2 * w/2) * 2== w/2){
			cout << "Yes" << endl;
		}else{
			cout << "No" << endl;
		}
	}else{
		if(four_count >= (h-1) / 2 * (w-1) / 2 && one_count == 1){
			cout << "Yes" << endl;
		}else{
			cout << "No" << endl;
		}
	}
	return 0;
}
int main() {
	// clock_t start = clock();
	cout << std::fixed << std::setprecision(10);
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
	ll end_solve = solve();
	/*
	clock_t end = clock();
	const double time = static_cast<double>(end - start) / CLOCKS_PER_SEC *
	1000.0; printf("time %lf[ms]\n", time);
	*/

	return (int)end_solve;
}