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"

#define REP(i,n) for(ll i=0;i<ll(n);++i)
#define RREP(i,n) for(ll i=ll(n)-1;i>=0;--i)
#define FOR(i,m,n) for(ll i=m;i<ll(n);++i)
#define RFOR(i,m,n) for(ll i=ll(n)-1;i>=ll(m);--i)
#define ALL(v) (v).begin(),(v).end()
#define UNIQUE(v) v.erase(unique(ALL(v)),v.end());
#define DUMP(v) REP(aa, (v).size()) { cout << v[aa]; if (aa != v.size() - 1)cout << " "; else cout << endl; }
#define INF 1000000001ll
#define MOD 1000000007ll
#define EPS 1e-9

const int dx[8] = { 1,1,0,-1,-1,-1,0,1 };
const int dy[8] = { 0,1,1,1,0,-1,-1,-1 };


using namespace std;
typedef long long ll;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector<vi> vvi;
typedef vector<vl> vvl;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
ll max(ll a, int b) { return max(a, ll(b)); }
ll max(int a, ll b) { return max(ll(a), b); }
ll min(ll a, int b) { return min(a, ll(b)); }
ll min(int a, ll b) { return min(ll(a), b); }
///(´・ω・`)(´・ω・`)(´・ω・`)(´・ω・`)(´・ω・`)(´・ω・`)///


int main() {
	cin.tie(0);
	ios::sync_with_stdio(false);
	int n, m;
	cin >> n >> m;
	vi cnt(26, 0);
	REP(i, n) {
		string s;
		cin >> s;
		REP(j, m)cnt[s[j] - 'a']++;
	}
	REP(i, 26)cnt[i] %= 4;
	int a = 0, b = 0, c = 0;
	REP(i, 26) {
		if (cnt[i] == 1)a++;
		else if (cnt[i] == 2)b++;
		else if (cnt[i] == 3)c++;
	}
	if (n % 2 ) {
		if (m % 2) {
			if ((a == 1 && c == 0&&b<=n/2+m/2) || (a == 0 && c == 1&&b<n/2+m/2))cout << "Yes" << endl;
			else cout << "No" << endl;
		}
		else {
			if (a == 0 && c == 0 && b <= m / 2)cout << "Yes" << endl;
			else cout << "No" << endl;
		}
	}
	else {
		if (m % 2) {
			if (a == 0 && c == 0 && b <= n / 2)cout << "Yes" << endl;
			else cout << "No" << endl;
		}
		else {
			if (a == 0 && b == 0 && c == 0)cout << "Yes" << endl;
			else cout << "No" << endl;
		}
	}
}