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;
 
template<class T> inline string toString(T x) {ostringstream sout;sout<<x;return sout.str();}

 
//typedef
//------------------------------------------
typedef pair<int, int> PII;
typedef pair<long, long> PLL;
typedef long long LL;

//container util
//------------------------------------------
#define PB emplace_back
#define MP make_pair 
#define SZ(a) int((a).size())
//repetition
//------------------------------------------
#define FOR(i,a,b) for(LL i=(a);i<(b);++i)
#define REP(i,n)  FOR(i,0,n)
#define SORT(c) sort((c).begin(),(c).end())
//constant
//--------------------------------------------
//clear memory
#define CLR(a) memset((a), 0 ,sizeof(a))
 
const LL INF=LONG_MAX;
const double EPS=0.00001;
const long double PI = acos(-1.0);

const int MAX_N=2e5+2;

int main(){
    int H,W;
    cin>>H>>W;
    int cnt[26];
    CLR(cnt);
    REP(h,H)REP(w,W){
        char c;
        cin>>c;
        cnt[c-'a']++;
    }
    vector<int>vec;
    int sum=0;
    REP(i,26){
        if(cnt[i]%4!=0){
            sum+=cnt[i]-(cnt[i]%4);
            vec.PB(cnt[i]%4);
        }else{
            sum+=cnt[i];
        }
    }
    bool f=1;
    if(H%2==0&&W%2==0){
        if(SZ(vec)!=0)f=0;
    }else if(H%2!=W%2){
        if(H%2==1&&sum<H*W-W)f=0;
        else if(W%2==1&&sum<H*W-H)f=0;
        REP(i,SZ(vec))if(vec[i]%2==1)f=0;
    }else{
        if(sum<H*W-(H+W-1))f=0;
        int odd=0;
        REP(i,SZ(vec)){
            if(vec[i]%2==1)odd++;
        }
        if(odd!=1)f=0;
    }
    if(f){
        cout<<"Yes"<<endl;
    }else{
        cout<<"No"<<endl;
    }
    return 0;
}