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)

//----------------------------おまじない
#pragma GCC optimize ("O3")
#pragma GCC target ("tune=native")
#pragma GCC target ("avx")
//----------------------------
#define FOR(i,j,n) for (int i=(j);i<(n);i++)
#define REP(i,n) for (int i=0;i<(n);i++)
#define REPN(i,n) for (int i=(n);i>=0;i--)
#define I(n) scanf("%d", &(n))
#define LL(n) scanf("%lld", &(n))
#define pb(n) push_back((n))
#define mp(i,j) make_pair((i),(j))
#define eb(i,j) emplace_back((i),(j))
#include <bits/stdc++.h>
using namespace std;

//------------------------------typedef集
typedef vector<int> vi;
typedef pair<int,int> pi;
typedef vector<pi> vpi;
typedef vector<vi> vvi;
typedef vector<vpi> vvpi;
typedef vector<vvi> vvvi;
typedef long long ll;
const int mod = 1000000009;

int h,w,cnt[26],two,one;
string s;

int main(){
    I(h); I(w);
    REP(i,h){
        cin >> s;
        REP(j,w) cnt[s[j]-'a']++;
    }
    if (h%2 && w%2){
        one = 1;
        two = h/2+w/2;
    } else if (h%2){
        two = w/2;
    } else if (w%2){
        two = h/2;
    }
    REP(i,26){
        if (cnt[i]&1) one--;
        if (cnt[i]&2) two--;
    }
    if (one >= 0 && two >= 0 ) {
        cout << "Yes" << endl;
    } else {
        cout << "No" << endl;
    }
}