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:
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <vector>
#include <map>
#include <set>
#include <bitset>
#include <queue>
#include <stack>
#include <sstream>
#include <cstring>
#include <numeric>
#include <ctime>
#include <cassert>
#include <random>
//#include <valarray>
//#include <fstream>

#define re return
#define fi first
#define se second
#define mp make_pair
#define pb emplace_back
#define all(x) x.begin(), x.end()
#define sz(x) ((int)(x).size())
#define rep(i, n) for (int i = 0; i < (n); i++)
#define rrep(i, n) for (int i = (n) - 1; i >= 0; i--)
#define y0 y32479
#define y1 y95874
#define next next239
#define prev prev239
#define fill(x, y) memset(x, y, sizeof(x))
#define sqr(x) ((x)*(x))
#define sqrt(x) sqrt(abs(x))
#define unq(x) (x.resize(unique(all(x)) - x.begin()))
#define ba back()
#define popcount __builtin_popcountll
#define firstbit(x) (63 - __builtin_clzll(x))
#define fastIO() {ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);}

#ifdef LOCAL_BOBER
bool local = true;
#define deb(x) cout << #x << " = " << (x) << endl
#define debn(x, n) { cout << #x << "(" << n << ") = " << \
	"{"; int _f_ = 1; rep(_i_, n) {if (!_f_) cout << "|"; cout << x[_i_]; _f_= 0;} cout << "}" << endl;}
#define deba(x) { cout << #x << " (size: " << sz(x) << ") = " << \
	"{"; int _f_ = 1; for (auto o : x) {if (!_f_) cout << "|"; cout << o; _f_ = 0;} cout << "}" << endl;}
#else
bool local = false;
#define deb(x) ;
#define debn(x, n) ;
#define deba(x) ;
#define endl "\n"
#endif

using namespace std;

typedef vector<int> vi;
typedef vector<vi> vvi;
typedef pair<int, int> ii;
typedef vector<ii> vii;
typedef vector<string> vs;
typedef long long ll;
typedef pair<ll, ll> pll;
typedef vector<ll> vll;
typedef pair<double, double> pdd;
typedef long double LD;
typedef double D;

template<class T> void print(T v) { cout << sz(v) << endl; for (auto x : v) cout << x << ' '; cout << endl; };
template<class T> void print1(T v) { cout << sz(v) << endl; for (auto x : v) cout << x + 1 << ' '; cout << endl; };
template<class T1, class T2> ostream& operator << (ostream &o, pair<T1, T2> x) {re o << x.fi << ' ' << x.se;}
template<class T1, class T2> istream& operator >> (istream &o, pair<T1, T2> &x) {re o >> x.fi >> x.se;}
template<class T> ostream& operator << (ostream &o, vector<T> &x) {for (auto &el : x) o << el << ' '; re o;}
template<class T> istream& operator >> (istream &o, vector<T> &x) {for (auto &el : x) o >> el; re o;}
template<class T1, class T2> pair<T1, T2> operator + (pair<T1, T2> a, pair<T1, T2> b) {a.fi += b.fi; a.se += b.se; re a;}
template<class T1, class T2> pair<T1, T2> operator - (pair<T1, T2> a, pair<T1, T2> b) {a.fi -= b.fi; a.se -= b.se; re a;}
template<class T1, class T2> void operator += (pair<T1, T2> &a, pair<T1, T2> b) {a.fi += b.fi; a.se += b.se;}
template<class T1, class T2> void operator -= (pair<T1, T2> &a, pair<T1, T2> b) {a.fi -= b.fi; a.se -= b.se;}

int nint() {int x; cin >> x; re x;}
double getTime() { re clock() / (double) CLOCKS_PER_SEC;};

mt19937 rnd(0);
int rand(int n) { re rnd() % n; }
int rand(int l, int r) { re rnd() % (r - l + 1) + l; }

const int mod = 1000000000 + 7;
//const int mod = 998244353;

void initIO() {
	if (local) {
		freopen("input.txt", "r", stdin);
		//freopen("output.txt", "w", stdout);
		srand((int)time(0));
		rnd.seed((int)time(0));
	}
	else {
		//freopen("input.txt", "r", stdin);
		//freopen("output.txt", "w", stdout);
		fastIO();
	}
}

void solve();
void precalc();

int TID;

signed main() {
	initIO();

	int tc = 1;
//	cin >> tc;

	precalc();
	rep(tt, tc) {
		TID = tt;
//		cout << "Case #" << tt + 1 << ": ";
		solve();
	}

	if (local)
		cout << endl << "time = " << getTime() << endl;
}

void precalc() {

}

/* ================ actual code starts here ================ */

int n;
int m;
string s;

int ne[20][200500];

int table[1 << 18];

int getans(int mask) {
	int ans = mod;
	rep(i, m) {
		if (mask & (1 << i)) {
			int tmp = table[mask ^ (1 << i)];
			if (tmp < n) {
				ans = min(ans, ne[i][tmp]);
			}
		}
	}
	re ans;
}

int check(int len) {
	int ne[20], last[20];
	rep(i, m)
		last[i] = n;
	rep(i, m)
		ne[i] = n + 1;
	rrep(p, n) {
		if (s[p] != '?')
			last[s[p] - 97] = p;
		rep(i, m) {
			int ok = 1;
			rep(j, m) {
				if (i != j && last[j] < p + len)
					ok = 0;
			}
			if (ok)
				ne[i] = p + len;
		}
		rep(i, m)
			::ne[i][p] = ne[i];
	}

	table[0] = 0;
	rep(i, 1 << m) {
		if (!i)
			continue;
		table[i] = getans(i);
	}

	re table[(1 << m) - 1] <= n;
}

void solve() {
	cin >> n >> m;
	cin >> s;

	int l = 1, r = n, ans = 0;
	while (l <= r) {
		int c = (l + r) / 2;
		if (check(c)) {
			ans = c;
			l = c + 1;
		}
		else
			r = c - 1;
	}
	cout << ans << endl;
}