Given is a string S of length N.

Find the maximum length of a non-empty string that occurs twice or more in S as contiguous substrings without overlapping.

More formally, find the maximum positive integer len such that there exist integers l_1 and l_2 ( 1 \leq l_1, l_2 \leq N - len + 1 ) that satisfy the following:

* l_1 + len \leq l_2

* S[l_1+i] = S[l_2+i] (i = 0, 1, ..., len - 1)




If there is no such integer len, print 0.

Constraints

* 2 \leq N \leq 5 \times 10^3
* |S| = N
* S consists of lowercase English letters.

Input

Input is given from Standard Input in the following format:


N
S


Output

Print the maximum length of a non-empty string that occurs twice or more in S as contiguous substrings without overlapping. If there is no such non-empty string, print 0 instead.

Examples

Input

5
ababa


Output

2


Input

2
xy


Output

0


Input

13
strangeorange


Output

5
import java.util.*;
import java.io.*;

public class Main {
	static int mod = (int) 1e9 + 7;
	static int DX[] = { -1, 0, 1, 0 }, DY[] = { 0, -1, 0, 1 };
	static final int INF = Integer.MAX_VALUE / 3;
	static final long LINF = Long.MAX_VALUE / 3;

	public static void main(String[] args) {
		FastScanner fs = new FastScanner(System.in);
		int n = fs.nextInt();
		String s = fs.next();
		char c[] = s.toCharArray();
		int r = 0;
		String now = "";
		int ans = 0;
		for(int l=0;l<n;l++) {
			if(now.length()>0)now = now.substring(1);
			while(l>r)r++;
			while(s.indexOf(now + c[r], r+1)>=0) {
				now += c[r];
				ans = Math.max(ans, r - l + 1);
				r++;
			}
		}
		System.out.println(ans);
	}

	//MOD culculations
	static long plus(long x, long y) {
		long res = (x + y) % mod;
		return res < 0 ? res + mod : res;
	}

	static long sub(long x, long y) {
		long res = (x - y) % mod;
		return res < 0 ? res + mod : res;
	}

	static long mul(long x, long y) {
		long res = (x * y) % mod;
		return res < 0 ? res + mod : res;
	}

	static long div(long x, long y) {
		long res = x * pow(y, mod - 2) % mod;
		return res < 0 ? res + mod : res;
	}

	static long pow(long x, long y) {
		if (y < 0)
			return 0;
		if (y == 0)
			return 1;
		if (y % 2 == 1)
			return (x * pow(x, y - 1)) % mod;
		long root = pow(x, y / 2);
		return root * root % mod;
	}
}

//高速なScanner
class FastScanner {
	private BufferedReader reader = null;
	private StringTokenizer tokenizer = null;

	public FastScanner(InputStream in) {
		reader = new BufferedReader(new InputStreamReader(in));
		tokenizer = null;
	}

	public String next() {
		if (tokenizer == null || !tokenizer.hasMoreTokens()) {
			try {
				tokenizer = new StringTokenizer(reader.readLine());
			} catch (IOException e) {
				throw new RuntimeException(e);
			}
		}
		return tokenizer.nextToken();
	}

	public String nextLine() {
		if (tokenizer == null || !tokenizer.hasMoreTokens()) {
			try {
				return reader.readLine();
			} catch (IOException e) {
				throw new RuntimeException(e);
			}
		}
		return tokenizer.nextToken("\n");
	}

	public long nextLong() {
		return Long.parseLong(next());
	}

	public int nextInt() {
		return Integer.parseInt(next());
	}

	public double nextDouble() {
		return Double.parseDouble(next());
	}

	public int[] nextIntArray(int n) {
		int[] a = new int[n];
		for (int i = 0; i < n; i++)
			a[i] = nextInt();
		return a;
	}

	public long[] nextLongArray(int n) {
		long[] a = new long[n];
		for (int i = 0; i < n; i++)
			a[i] = nextLong();
		return a;
	}
}