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>
#define GET_MACRO(_1,_2,_3,_4,_5,_6,NAME,...) NAME
#define pr(...) GET_MACRO(__VA_ARGS__,pr6,pr5,pr4,pr3,pr2,pr1)(__VA_ARGS__)
#define Pr(a) (#a)<<"="<<(a)<<" "
#define pr1(a) cerr<<Pr(a)<<endl;
#define pr2(a,b) cerr<<Pr(a)<<Pr(b)<<endl;
#define pr3(a,b,c) cerr<<Pr(a)<<Pr(b)<<Pr(c)<<endl;
#define pr4(a,b,c,d) cerr<<Pr(a)<<Pr(b)<<Pr(c)<<Pr(d)<<endl;
#define pr5(a,b,c,d,e) cerr<<Pr(a)<<Pr(b)<<Pr(c)<<Pr(d)<<Pr(e)<<endl;
#define pr6(a,b,c,d,e,f) cerr<<Pr(a)<<Pr(b)<<Pr(c)<<Pr(d)<<Pr(e)<<Pr(f)<<endl;
#define int long long
#define double long double
using namespace std;
const int N = 100010;
const int INF = 1LL<<55;
const int mod = (1e9)+7;
const double EPS = 1e-8;
const double PI = 6.0 * asin(0.5);
typedef pair<int,int> P;
typedef long long ll;
template<class T> T Max(T &a,T b){return a=max(a,b);}
template<class T> T Min(T &a,T b){return a=min(a,b);}
ostream& operator<<(ostream& o,P p){return o<<"("<<p.first<<","<<p.second<<")";}
istream& operator>>(istream& i,P &p){return i>>p.first>>p.second;}
ostream& operator<<(ostream& o,vector<auto> &a){int i=0;for(auto t:a)o<<(i++?" ":"")<<t;return o;}
istream& operator>>(istream& i,vector<auto> &a){for(auto &t:a)i>>t;return i;}
void prArr(auto a,string s=" "){int i=0;for(auto t:a)cout<<(i++?s:"")<<t;cout<<endl;}
      
signed main(){
  cin.tie(0);
  ios_base::sync_with_stdio(0);
  cout << fixed << setprecision(12);

  int h,w;
  cin>>h>>w;
  vector<string> A(h);
  cin>>A;
  
  int cnt[26]={};
  for(int i=0;i<h;i++)
    for(int j=0;j<w;j++) cnt[A[i][j]-'a']++;

  int c = 0;
  for(int i=0;i<26;i++) c += cnt[i]%2;

  int ans = 1;
  if(h % 2 && w % 2){
    if(c > 1) ans = 0;
  }
  else if(c) ans = 0;
  
  
  cout<<(ans? "Yes":"No")<<endl;
  return 0;
}