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)

#include "bits/stdc++.h"
using namespace std;
#define ll long long int
#define rep(i,n) for( int i = 0; i < n; i++ )
#define rrep(i,n) for( int i = n; i >= 0; i-- )
#define REP(i,s,t) for( int i = s; i <= t; i++ )
#define RREP(i,s,t) for( int i = s; i >= t; i-- )
#define dump(x)  cerr << #x << " = " << (x) << endl;
#define INF 2000000000
#define mod 1000000007
#define INF2 1000000000000000000


int main(void)
{
    cin.tie(0);
    ios::sync_with_stdio(false);
    int H, W; cin >> H >> W;
    string S[110];
    rep(i, H) cin >> S[i];
    int cnt[26] = {};
    rep(i, H) rep(j, W) cnt[S[i][j] - 'a']++;
    sort(cnt, cnt + 26);
    int k = 0;
    int used1 = 0, used2 = 0, used4 = 0;
    rep(i, (H + 1) / 2) rep(j, (W + 1) / 2) {
        if (i < H / 2 && j < W / 2) used4++;
        else if (i == H / 2 && j < W / 2) used2++;
        else if (i < H / 2 && j == W / 2) used2++;
        else used1++;
    }
    int cnt1 = 0, cnt2 = 0, cnt4 = 0;
    rep(i, 26) {
        if (cnt[i] % 4 == 1) {cnt1++; cnt[i]--;}
        if (cnt[i] % 4 == 2) {cnt2++; cnt[i] -= 2;}
        cnt4 += cnt[i] / 4; 
    }
    int less = min(used1, cnt1);
    used1 -= less;
    cnt1 -= less;
    used2 += used1 / 2;
    used1 -= (used1 / 2) * 2;
    less = min(used2, cnt2);
    used2 -= less;
    cnt2 -= less;
    used4 += used2 / 2;
    used2 -= (used2 / 2) * 2;
    less = min(used4, cnt4);
    used4 -= less;
    cnt4 -= less;
    if(used1 > 0 || used2 > 0 || used4 > 0) {
        cout << "No" << endl;
    }else {
        cout << "Yes" << endl;
    }
    return 0;
}