Problem:
Score : 100 points

Problem Statement

Kyoto University Programming Contest is a programming contest voluntarily held by some Kyoto University students.
This contest is abbreviated as Kyoto University Programming Contest and called KUPC.

source: Kyoto University Programming Contest Information

The problem-preparing committee met to hold this year's KUPC and N problems were proposed there.
The problems are numbered from 1 to N and the name of i-th problem is P_i.
However, since they proposed too many problems, they decided to divide them into some sets for several contests.

They decided to divide this year's KUPC into several KUPCs by dividing problems under the following conditions.

One KUPC provides K problems.

Each problem appears at most once among all the KUPCs.

All the first letters of the problem names in one KUPC must be different.

You, one of the committee members, want to hold as many KUPCs as possible.
Write a program to find the maximum number of KUPCs that can be held this year.

Constraints

1 \leq N \leq 10^4

1 \leq K \leq 26

1 \leq |P_i| \leq 10

All characters in P_i are capital letters.

Note that, for each i and j (1 \leq i < j \leq N),
P_i \neq P_j are not necessarily satisfied.

Input

The input is given from Standard Input in the following format:

N K
P_1
:
P_N

Output

Print the maximum number of KUPCs that can be held on one line.

Sample Input 1

9 3
APPLE
ANT
ATCODER
BLOCK
BULL
BOSS
CAT
DOG
EGG

For example, three KUPCs can be held by dividing the problems as follows.

First: APPLE, BLOCK, CAT

Second: ANT, BULL, DOG

Third: ATCODER, BOSS, EGG

Sample Output 1

3

Sample Input 2

3 2
KU
KYOUDAI
KYOTOUNIV

Sample Output 2

0

No KUPC can be held.

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;

using vi = vector<int>; using vvi = vector<vi>; using vvvi = vector<vvi>;
using ll = long long int;
using vll = vector<ll>; using vvll = vector<vll>; using vvvll = vector<vvll>;
using vd = vector<double>; using vvd = vector<vd>; using vvvd = vector<vvd>;
using P = pair<int, int>;
using Pll = pair<ll, ll>;
using cdouble = complex<double>;

const double eps = 1e-9;
const double INFD = numeric_limits<double>::infinity();
const double PI = 3.14159265358979323846;
#define Loop(i, n) for(int i = 0; i < (int)(n); i++)
#define Loop1(i, n) for(int i = 1; i <= (int)(n); i++)
#define Loopr(i, n) for(int i = (int)(n) - 1; i >= 0; i--)
#define Loopr1(i, n) for(int i = (int)(n); i >= 1; i--)
#define Foreach(buf, container) for(auto buf : container)
#define Loopdiag(i, j, h, w, sum) for(int i = ((sum) >= (h) ? (h) - 1 : (sum)), j = (sum) - i; i >= 0 && j < (w); i--, j++)
#define Loopdiagr(i, j, h, w, sum) for(int j = ((sum) >= (w) ? (w) - 1 : (sum)), i = (sum) - j; j >= 0 && i < (h); j--, i++)
#define Loopdiagsym(i, j, h, w, gap) for (int i = ((gap) >= 0 ? (gap) : 0), j = i - (gap); i < (h) && j < (w); i++, j++)
#define Loopdiagsymr(i, j, h, w, gap) for (int i = ((gap) > (h) - (w) - 1 ? (h) - 1 : (w) - 1 + (gap)), j = i - (gap); i >= 0 && j >= 0; i--, j--)
#define Loopitr(itr, container) for(auto itr = container.begin(); itr != container.end(); itr++)
#define printv(vector) Loop(ex_i, vector.size()) { cout << vector[ex_i] << " "; } cout << endl;
#define printmx(matrix) Loop(ex_i, matrix.size()) { Loop(ex_j, matrix[ex_i].size()) { cout << matrix[ex_i][ex_j] << " "; } cout << endl; }
#define quickio() ios::sync_with_stdio(false); cin.tie(0);
#define bitmanip(m,val) static_cast<bitset<(int)m>>(val)
#define Comp(type_t) bool operator<(const type_t &another) const
#define fst first
#define snd second
bool nearlyeq(double x, double y) { return abs(x - y) < eps; }
bool inrange(int x, int t) { return x >= 0 && x < t; }
bool inrange(vi xs, int t) { Foreach(x, xs) if (!(x >= 0 && x < t)) return false; return true; }
ll rndf(double x) { return (ll)(x + (x >= 0 ? 0.5 : -0.5)); }
ll floorsqrt(ll x) { ll m = (ll)sqrt((double)x); return m + (m * m <= x ? 0 : -1); }
ll ceilsqrt(ll x) { ll m = (ll)sqrt((double)x); return m + (x <= m * m ? 0 : 1); }
ll rnddiv(ll a, ll b) { return (a / b + (a % b * 2 >= b ? 1 : 0)); }
ll ceildiv(ll a, ll b) { return (a / b + (a % b == 0 ? 0 : 1)); }
ll gcd(ll m, ll n) { if (n == 0) return m; else return gcd(n, m % n); }
ll lcm(ll m, ll n) { return m * n / gcd(m, n); }

/*******************************************************/

// range = [l, r), return last value causing "t" in evalfunc that returns l->[t,...,t,f,...,f)->r
// NOTE: if [f,...,f) then return l - 1, if [l, r) = empty set then invalid use
template<typename val_t, typename bsargv_t>
val_t lower_binary_search(val_t l, val_t r, const bsargv_t &v, bool(*evalfunc)(val_t, const bsargv_t&)) {
	if (r - l == 1) {
		if (evalfunc(l, v)) return l;
		else return l - 1;
	}
	val_t m = (l + r) / 2;
	if (evalfunc(m, v)) return lower_binary_search<val_t, bsargv_t>(m, r, v, evalfunc);
	else return lower_binary_search<val_t, bsargv_t>(l, m, v, evalfunc);
}

struct bsargv_t {
	vi cnt;
	int k;
};

bool evalfunc(int val, const bsargv_t &v) {
	priority_queue<int, vi, greater<int>> pq;
	Loop(i, val) pq.push(0);
	vi buf;
	Loop(i, 26) {
		Loop(j, min(val, v.cnt[i])) {
			int x = pq.top(); pq.pop();
			buf.push_back(x + 1);
		}
		while (buf.size()) {
			pq.push(buf.back());
			buf.pop_back();
		}
	}
	if (pq.size() && pq.top() >= v.k) return true;
	else return false;
}

int main() {
	int n, k; cin >> n >> k;
	vi cnt(26);
	Loop(i, n) {
		string s; cin >> s;
		cnt[s[0] - 'A']++;
	}
	bsargv_t v = { cnt, k };
	cout << lower_binary_search<int>(1, n + 1, v, evalfunc) << endl;
}