Problem: You are given a string s of length n. Each character is either one of the first k lowercase Latin letters or a question mark.

You are asked to replace every question mark with one of the first k lowercase Latin letters in such a way that the following value is maximized.

Let f_i be the maximum length substring of string s, which consists entirely of the i-th Latin letter. A substring of a string is a contiguous subsequence of that string. If the i-th letter doesn't appear in a string, then f_i is equal to 0.

The value of a string s is the minimum value among f_i for all i from 1 to k.

What is the maximum value the string can have?

Input

The first line contains two integers n and k (1 ≤ n ≤ 2 ⋅ 10^5; 1 ≤ k ≤ 17) — the length of the string and the number of first Latin letters used.

The second line contains a string s, consisting of n characters. Each character is either one of the first k lowercase Latin letters or a question mark.

Output

Print a single integer — the maximum value of the string after every question mark is replaced with one of the first k lowercase Latin letters.

Examples

Input


10 2
a??ab????b


Output


4


Input


9 4
?????????


Output


2


Input


2 3
??


Output


0


Input


15 3
??b?babbc??b?aa


Output


3


Input


4 4
cabd


Output


1

Note

In the first example the question marks can be replaced in the following way: "aaaababbbb". f_1 = 4, f_2 = 4, thus the answer is 4. Replacing it like this is also possible: "aaaabbbbbb". That way f_1 = 4, f_2 = 6, however, the minimum of them is still 4.

In the second example one of the possible strings is "aabbccdda".

In the third example at least one letter won't appear in the string, thus, the minimum of values f_i is always 0.

Solution (in CPP):

Solution:
// Template Headers {{{
#include <bits/stdc++.h>
using namespace std;

using vi = vector<int>; using vvi = vector<vi>; using pii = pair<int, int>; using vpii = vector<pii>;
using ll = long long; using vll = vector<long long>; using vvll = vector<vll>; using pll = pair<ll, ll>; using vpll = vector<pll>;
template <class T> using min_queue = priority_queue<T, vector<T>, greater<T>>;
template <class T> istream &operator>>(istream &, vector<T> &); template <class T> ostream &operator<<(ostream &, const vector<T> &);
template <class T, class U> istream &operator>>(istream &, pair<T, U> &); template <class T, class U> ostream &operator<<(ostream &, const pair<T, U> &);
template <class T = int> constexpr T INFTY = numeric_limits<T>::has_infinity ? numeric_limits<T>::infinity() : numeric_limits<T>::max();
template <> constexpr int INFTY<int> = 0x3f3f3f3f; template <> constexpr ll INFTY<ll> = 0x3f3f3f3f3f3f3f3fLL;
template <class T, class U> constexpr pair<T, U> INFTY<pair<T, U>> = make_pair(INFTY<T>, INFTY<U>);
constexpr int INF = INFTY<>; constexpr ll BINF = INFTY<ll>;
// mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
// }}}
// Preprocessing & "Global" variables {{{
struct Solver {
	Solver() {}
	int solve();
};
// }}}

template <class T, class Predicate>
T getFirstTrue(Predicate f, T lo = 0, T hi = INFTY<T>) {
	while (lo < hi) {
		int mi = lo + (hi - lo) / 2;
		if (f(mi))
			hi = mi;
		else
			lo = mi + 1;
	}
	return lo;
}
template <class T, class Predicate>
T getLastTrue(Predicate f, T lo = 0, T hi = INFTY<T>) {
	while (lo < hi) {
		int mi = lo + (hi - lo) / 2;
		if (!f(mi))
			hi = mi;
		else
			lo = mi + 1;
	}
	return lo - 1;
}


int Solver::solve() {
	int n, k; cin >> n >> k;
	string s; cin >> s;
	vvi mxl(k, vi(n));
	for (int i = 0; i < n; i++) {
		for (int j = 0; j < k; j++) {
			if (s[i] == 'a' + j || s[i] == '?') {
				mxl[j][i] = 1;
				if (i > 0) mxl[j][i] += mxl[j][i - 1];
			}
		}
	}
	auto solve = [&] (int l) {
		if (l == 0) return true;
		vvi lm_ok(k, vi(n, INF)); // left-most ok position starting from i
		for (int i = n - 1; i >= 0; i--) {
			for (int j = 0; j < k; j++) {
				if (i < n - 1) lm_ok[j][i] = lm_ok[j][i + 1];
				if (mxl[j][i] >= l) lm_ok[j][i] = i;
			}
		}
		vi dp(1 << k, INF);
		dp[0] = -1;
		for (int msk = 0; msk < 1 << k; msk++) {
			if (dp[msk] >= INF) continue;
			int can = dp[msk] + l;
			if (can >= n) continue;
			for (int j = 0; j < k; j++) {
				int nmsk = msk | (1 << j);
				if (nmsk != msk) {
					dp[nmsk] = min(dp[nmsk], lm_ok[j][can]);
				}
			}
		}
		return dp.back() < n;
	};
	cout << getLastTrue(solve, 0, n + 1) << endl;
	return 0;
}

// Template Main {{{
int main() {
	ios::sync_with_stdio(0); cin.tie(0);
	Solver solver;
	#ifdef JUDGE_MULTI_CASE
		int t; cin >> t; for (int i = 1; i <= t; i++)
	#endif
	solver.solve();
	// cout << "Case #" << i << ": ", solver.solve();
	return 0;
}

template <class T> istream &operator>>(istream &is, vector<T> &v) {
	for (auto it=v.begin();it!=v.end();++it) is >> *it;
	return is;
}

template <class T> ostream &operator<<(ostream &os, const vector<T> &v) {
	for (auto it=v.begin();it!=v.end();) os << *it, os << " \n"[++it==v.end()];
	return os;
}

template <class T, class U> istream &operator>>(istream &is, pair<T, U> &p) {
	return is >> p.first >> p.second;
}

template <class T, class U> ostream &operator<<(ostream &os, const pair<T, U> &p) {
	return os << p.first << ' ' << p.second;
}
// }}}

// vim: fdm=marker