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 pb push_back
#define mp make_pair
#define forn(i, n) for (int i = 0; i < (int)(n); ++i)
typedef long long LL;
typedef pair<int, int> PII;

int n, m;
char s[100][101];
bool used[100][100];
int sz[5];
int cnt[26];

int main() {
    scanf("%d%d", &n, &m);
    forn(i, n) scanf("%s", s[i]);
    forn(i, n) forn(j, m) {
        ++cnt[s[i][j] - 'a'];
    }
    forn(i, n) forn(j, m) {
        if (used[i][j]) {
            continue;
        }
        vector<PII> v;
        forn(p, 2) forn(q, 2) {
            int ni = (p ? i : (n - 1 - i));
            int nj = (q ? j : (m - 1 - j));
            if (used[ni][nj]) {
                continue;
            }
            used[ni][nj] = true;
            v.pb(mp(ni, nj));
        }
        ++sz[v.size()];
    }
    for (int i = 4; i >= 1; --i) {
        forn(_, sz[i]) {
            bool ok = false;
            forn(j, 26) if (cnt[j] >= i) {
                cnt[j] -= i;
                ok = true;
                break;
            }
            if (!ok) {
                cout << "No" << endl;
                return 0;
            }
        }
    }
    cout << "Yes" << endl;
    return 0;
}