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 PI 3.14159265358979323846
#define MAXINF (1e18L)
#define INF (1e9L)
#define EPS (1e-9)
#define MOD ((ll)(1e9+7))
#define REP(i, n) for(int i=0;i<int(n);++i)
#define Rep(i,sta,n) for(int i=sta;i<n;i++)
#define RREP(i, n) for(int i=int(n)-1;i>=0;--i)
#define ALL(v) v.begin(),v.end()
#define FIND(v,x) (binary_search(ALL(v),(x)))
#define SORT(v) sort(ALL(v))
#define RSORT(v) sort(ALL(v));reverse(ALL(v))
#define DEBUG(x) cerr<<#x<<": "<<x<<endl;
#define DEBUG_VEC(v) cerr<<#v<<":";for(int i=0;i<v.size();i++) cerr<<" "<<v[i]; cerr<<endl
#define Yes(n) cout<<((n)?"Yes":"No")<<endl
#define YES(n) cout<<((n)?"YES":"NO")<<endl
#define pb push_back
#define fi first
#define se second
using namespace std;
template<class A>void pr(A a){cout << (a) << endl;}
template<class A,class B>void pr(A a,B b){cout << a << " "  ;pr(b);}
template<class A,class B,class C>void pr(A a,B b,C c){cout << a << " " ;pr(b,c);}
template<class A,class B,class C,class D>void pr(A a,B b,C c,D d){cout << a << " " ;pr(b,c,d);}
template<class T> void pr_vec(vector<T> &v, char ep=' '){int n=v.size();REP(ni,n){cout<<v[ni];if(ni!=n-1)cout<<ep;}cout<<endl;};
template<class T> inline bool chmin(T& a, T b){return a>b ? a=b, true : false;}
template<class T> inline bool chmax(T& a, T b){return a<b ? a=b, true : false;}
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll,ll> pll;

int main(void)
{
    int H,W;
    cin >> H >> W;
    map<int, int> m;
    for (int i = 0; i < H; i++) {
        string s;
        cin >> s;
        for(auto&& si: s) m[si-'a']++;
    }
    int t2=0, t1=0, t0=0;
    for(auto&& mi : m){
        t0 += mi.se / 4;
        if((mi.se%4)==0) continue;

        if((mi.se%2)==1){
            t1++;
        }else if((mi.se%2)==0){
            t2++;
        }
    }
    int t = t0+t1+t2;

    if((H%2)==0&&(W%2)==0){
        Yes(t0==t);
    }else if((H%2)==1&&(W%2)==1){
        Yes(t1==1&&t2==((H/2)+(W/2)));
    }else{
        if(H%2==1) Yes((t2+t0)==t&&t2==(W/2));
        if(W%2==1) Yes((t2+t0)==t&&t2==(H/2));
    }
}