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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;
int H,W;
int No() { return puts("No") * 0;}
string paindram(vector<char>even, vector<char>odd){
    string res(W,'x');
    assert(W % 2 == odd.size());
    if(W % 2 == 1){
        res[W / 2] = odd.front();
    }
    for(int i = 0; i < even.size(); ++i)
        res[i] = res[W - 1 - i] = even[i];
    return res;
}
int main(){
    cin >> H >> W;
    vector<string>g(H);
    vector<string>ans(H,string(W,'z'));
    for(auto& s : g) cin >> s;
    vector<vector<char>>even(H);
    vector<vector<char>> odd(H);
    int idx  = 0;
    for(auto s : g){
        map<char,int>cnt;
        for(auto c :s) cnt[c] += 1;
        for(auto [c,num] : cnt){
            if(num % 2){
                odd[idx].push_back(c);
            }else{
                num /= 2;
                while(num > 0){
                    even[idx].push_back(c);
                    num /= 2;
                }
            }
        }
        if(W % 2 == 0){
            if(odd[idx].size() >= 1)
                return No();
        } else {
            if(odd[idx].size() != 1)
                return No();
        }
        ++idx;
    }
    //we can construct paindram sentence,
    int remu = 0;
    vector<bool>used(H,false);
    for(int i = 0; i < even.size(); ++i){
        if(used[i]) continue;
        for(int j = 0; j < even.size(); ++j){
            if(i == j or used[j])continue;
            if(even[i] == even[j] && odd[i] == odd[j]){
                used[i] = used[j] = true;
                ans[remu] = paindram(even[i],odd[j]);
                ans[H - remu - 1] = paindram(even[i],odd[i]);
                remu++;
                break;
            }
        }
    }
    int unused = count_if(used.begin(), used.end(),[](bool e){
        return !e;
    });
    if(H % 2 == 0 && unused > 0) return No();
    if(H % 2 == 1 && unused != 1) return No();
    if(H % 2 == 1){
        auto itr = find(used.begin(), used.end(),false);
        int id = distance(used.begin(),itr);
        ans[H / 2] = paindram(even[id],odd[id]);
    }
    cout << "Yes" << endl;
}