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<string>
#include<vector>
#include<algorithm>
#include<map>
#include<set>
#include<cstdio>
#include<cmath>
#include<numeric>
#include<queue>
#include<stack>
#include<cstring>
#include<limits>
#include<functional>
#define rep(i,a) for(int i=(int)0;i<(int)a;++i)
#define pb push_back
#define eb emplace_back
using ll=long long;
static const ll mod = 1e9 + 7;
static const ll INF = 1LL << 50;
using namespace std;

int main(){
    int h,w;
    cin>>h>>w;
    map<char,int>mp;
    rep(i,h)rep(j,w){
        char s;
        cin>>s;
        mp[s]++;
    }
    int four=max(h/2,w/2);
    if(!(h/2&&w/2))four=0;
    int a=four;
    if(four)for(auto x:mp){
        if(x.second>=4){
            four-=x.second/4;
            mp[x.first]-=4*(x.second/4);
        }
    }
    int two=(h*w-4*a)/2;
    int b=two;
    if(two)for(auto x:mp){
        if(x.second>=2){
            two-=x.second/2;
            mp[x.first]-=2*x.second/2;
        }
    }
    int one=(h*w-4*a-2*b);
    if(one)for(auto x:mp){
        if(x.second>=1){
            one-=x.second;
            mp[x.first]-=x.second;
        }
    }
    if(one!=0||two!=0||four!=0)cout<<"No"<<endl;
    else cout<<"Yes"<<endl;
    return 0;
    }