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>

using namespace std;

#define int long long

using ll = long long;
using pii =  pair<int, int>;

const int IINF =  0x3f3f3f3f;
const ll LINF = 0x3f3f3f3f3f3f3f3fLL;
const int INF = sizeof(int) == sizeof(long long) ? LINF : IINF;
const ll MOD = 1e9+7;

#define for_(i,a,b) for(int i = (a);i < (b);i++)
#define rfor_(i,a,b) for(int i = (b)-1;i >= (a);i--)

#define rep(i,n) for_(i,0,n)
#define rrep(i,n) rfor_(i,0,n)

#define pb push_back
//#define eb emplace_back
#define mp make_pair
#define ft first
#define sd second

#define all(a) (a).begin(), (a).end()
#define sz(a) (int)(a).size()
#define rev(a) reverse(all(a))

#define endl '\n'

const int limit = 100007;

int H, W;

char c[104][104];

char a[27];
int t, u;

signed main(){
	cin.tie(0);
	cout.tie(0);
	ios_base::sync_with_stdio(false);

	cin>>H>>W;

	rep(i, H)cin>>c[i];

	rep(i, H) {
		rep(j, W)a[c[i][j]-'a']++;
	}

	if(H%2==0 && W%2==0) {
		rep(i, 26) {
			if(a[i]%4!=0) {
				cout<<"No"<<endl;
				return 0;
			}
		}
		cout<<"Yes"<<endl;
	}else if(H%2==0 || W%2==0) {
		rep(i, 26) {
			if(a[i]%2==1) {
				cout<<"No"<<endl;
				return 0;
			}else if(a[i]%4==2)u++;
		}
		if((H%2==0 && 2*u<=H) && (H-2*u)%4==0)cout<<"Yes"<<endl;
		else if((W%2==0 && 2*u<=W) && (W-2*u)%4==0)cout<<"Yes"<<endl;
		else cout<<"No"<<endl;
	}else {
		rep(i, 26) {
			if(a[i]%2==1) {
				if(t==0) {
					t++;
					if(a[i]%4==3)u++;
				}else {
					cout<<"No"<<endl;
					return 0;
				}
			}else if(a[i]%4==2)u++;
		}
		if(H==1 || W==1)cout<<"Yes"<<endl;
		else if(2*u<=(H+W-2) && (H+W-2-2*u)%4==0)cout<<"Yes"<<endl;
		else cout<<"No"<<endl;
	}

		return 0;
	}