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>
#define rep(i, a, b) for(int i = a; i < b; i++)
#define Rep(i, a, b) for(int i = a; i <= b; i++)
#define repr(i, a, b) for(int i = b-1; i >= a; i--)
#define _GLIBCXX_DEBUG
#define Vl vector<ll>
#define Vs vector<string>
#define Vp vector<pair<ll, ll>>
using ll = long long;
#define ALL(v) (v).begin(),(v).end()
#define endl "\n"
#define chmin(x, y) x = min(x, y)
#define chmax(x, y) x = max(x, y)
#define co(x) cout << x << endl
#define coel cout << endl
#define pb push_back
#define sz(v) ((ll)(v).size())
using namespace std;  
const double pi = acos(-1.0);
const ll MOD = 1e9 + 7;
const ll INF = 1LL << 60;
#define pp pair<ll, pair<ll, ll>> 
#define fi first
#define se second

////////////////////////////////////////////////////////

ll b[30];

int main() {
    ll h, w; cin >> h >> w;
    rep(i, 0, h){
        string s; cin >> s;
        rep(j, 0, w){
            b[s[j]-'a']++;
        }
    }
    
    ll g1 = 0, g2 = 0, g4 = 0;
    
    if(h%2 != 0 && w%2 != 0){
        g1 = 1;
        g2 = h/2 + w/2;
        g4 = (h/2)*(w/2);
    }else{
        g1 = 0;
        g4 = (h/2)*(w/2);
        g2 = (h*w - g4*4)/2;
    }

    // cout << "g1 : " << g1 << endl;
    // cout << "g2 : " << g2 << endl;
    // cout << "g4 : " << g4 << endl;

    // rep(i, 0, 25) cout << b[i] << " \n"[i==24];

    if(g1){
        rep(i, 0, 25){
            if(b[i] == 1){
                b[i]--;
                break;
            }
        }
    } 
    ll t = 0;
    while(g4){
        rep(i, 0, 25){
            if(b[i] >= 4){
                b[i] -= 4;
                g4--;
            }
        }
        t++;
        if(t >= 100000){
            co("No");
            return 0;
        }
    }
    
    // rep(i, 0, 25) cout << b[i] << " \n"[i==24];

    t = 0;
    while(g2){
        rep(i, 0, 25){
            if(b[i] && b[i]%2 == 0){
                b[i] -= 2;
                g2--;
            }
        }
        t++;
        if(t >= 100000){
            co("No");
            return 0;
        }
    }

    // rep(i, 0, 25) cout << b[i] << " \n"[i==24];

    bool f = true;
    rep(i, 0, 25){
        if(b[i]) f = false;
    }
    if(f) co("Yes");
    else co("No");

    return 0;
}