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>
#include <vector>
#include <cmath>
#define REP(i, n) for (int i = 0; i < (int)(n); ++i)
#define REP2(i, m, n) for (int i = (m); i < (int)(n); ++i)
#define REPR(i, n) for (int i = (n)-1; i >= 0; --i)
#define REPR2(i, m, n) for (int i = (n)-1; i >= (m); --i)
#define REPx(x, a) for(auto x : a)
#define ALL(a) a.begin(), a.end()
#define SORT(a) sort(ALL(a))
#define SORTG(a) sort(ALL(a), greater<int>())
#define REVERSE(a) reverse(ALL(a))
#define bit_search(bit, n) REP(bit, 1<<(n))
#define bit_check(bit, i) (bit>>(i)) & 1
#define setpre(n) fixed << setprecision((n))
#define UNIQUE(a) do {SORT(a); (a).erase(unique(ALL(a)), (a).end());} while(0)
#define MAX(a) *max_element(ALL(a))
#define MIN(a) *min_element(ALL(a))
#define bisect_left(a, x) lower_bound(ALL(a), (x)) - a.begin()
#define bisect_right(a, x) upper_bound(ALL(a), (x)) - a.begin()
#define INPUT(a) REP(i, a.size()) cin >> a[i];
#define INPUTP(a) REP(i, a.size()) cin >> a[i].first >> a[i].second;
#define ENDL cout << endl;
using namespace std;
using ll = long long;
using ld = long double;
using PII = pair<int, int>;
using VPII = vector<PII>;
using VI = vector<int>;
using VVI = vector<VI>;
using Vll = vector<ll>;
using VVll = vector<Vll>;
using VC = vector<char>;
using VS = vector<string>;
using VB = vector<bool>;
const int INF = 1e9;
const ll MOD = 1e9 + 7;

int main()
{
  cin.tie(0);
  ios::sync_with_stdio(false);
  int H, W;
  cin >> H >> W;
  map<int, int> cnt;
  int g1, g2, g4;
  if (H % 2 && W % 2) {g1 = 1; g2 = (H+W-1)/2; g4 = (H*W-(g1+2*g2))/4;}
  else if (H % 2 == 0 && W % 2 == 0) {g1 = 0; g2 = 0; g4 = H*W/4;}
  else if (H % 2 && W % 2 == 0) {g1 = 0; g2 = W/2; g4 = (H*W-2*g2)/4;}
  else if (H % 2 == 0 && W % 2) {g1 = 0; g2 = H/2; g4 = (H*W-2*g2)/4;}
  REP(i, H) REP(j, W)
  {
    char a;
    cin >> a;
    cnt[a]++;
  }

  REP(i, g1) REPx(x, cnt)
  {
    char s = x.first;
    int ct = x.second;
    if (ct == 0) continue;
    if (ct % 4 == 1 || ct % 4 == 3)
    {
      cnt[s]--;
      break;
    }
  }
  REP(i, g2) REPx(x, cnt)
  {
    char s = x.first;
    int ct = x.second;
    if (ct == 0) continue;
    if (ct % 4 == 2)
    {
      cnt[s] -= 2;
      break;
    }
  }
  REP(i, g4) REPx(x, cnt)
  {
    char s = x.first;
    int ct = x.second;
    if (ct == 0) continue;
    if (ct % 4 == 0)
    {
      cnt[s] -= 4;
      break;
    }
  }
  REPx(x, cnt) if (x.second)
  {
    cout << "No" << endl;
    return 0;
  }
  cout << "Yes" << endl;

  return 0;
}