When Xellos was doing a practice course in university, he once had to measure the intensity of an effect that slowly approached equilibrium. A good way to determine the equilibrium intensity would be choosing a sufficiently large number of consecutive data points that seems as constant as possible and taking their average. Of course, with the usual sizes of data, it's nothing challenging — but why not make a similar programming contest problem while we're at it?

You're given a sequence of n data points a1, ..., an. There aren't any big jumps between consecutive data points — for each 1 ≤ i < n, it's guaranteed that |ai + 1 - ai| ≤ 1.

A range [l, r] of data points is said to be almost constant if the difference between the largest and the smallest value in that range is at most 1. Formally, let M be the maximum and m the minimum value of ai for l ≤ i ≤ r; the range [l, r] is almost constant if M - m ≤ 1.

Find the length of the longest almost constant range.

Input

The first line of the input contains a single integer n (2 ≤ n ≤ 100 000) — the number of data points.

The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 100 000).

Output

Print a single number — the maximum length of an almost constant range of the given sequence.

Examples

Input

5
1 2 3 3 2


Output

4


Input

11
5 4 5 5 6 7 8 8 8 7 6


Output

5

Note

In the first sample, the longest almost constant range is [2, 5]; its length (the number of data points in it) is 4.

In the second sample, there are three almost constant ranges of length 4: [1, 4], [6, 9] and [7, 10]; the only almost constant range of the maximum length 5 is [6, 10].
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.util.HashMap;
import java.util.Map;
import java.util.StringTokenizer;

public class P2 {

  static BufferedReader in;
  static PrintWriter out;
  static StringTokenizer tok;

  static void solve() throws Exception {
    int n = nextInt();
    int[] a = nextIntArray(n, 0);
    out.println(helper(a));
  }

  static int helper(int[] a) {
    int n = a.length;
    int k = 2;
    int ans = 0;
    int i = 0, j = 0;
    Map<Integer, Integer> map = new HashMap<>();
    while (j < n) {
      int c = a[j];
      if (map.containsKey(c)) {
        map.put(c, map.get(c) + 1);
      } else {
        while (map.size() == k) {
          int old = a[i];
          int count = map.get(old);
          if (count == 1) {
            map.remove(old);
          } else {
            map.put(old, count - 1);
          }
          i++;
        }
        map.put(c, 1);
      }
      ans = Math.max(ans, j - i + 1);
      j++;
    }
    return ans;
  }

  public static void main(String args[]) {
    try {
      in = new BufferedReader(new InputStreamReader(System.in));
      out = new PrintWriter(new OutputStreamWriter(System.out));
      solve();
      in.close();
      out.close();
    } catch (Throwable e) {
      e.printStackTrace();
      System.exit(1);
    }
  }

  static int nextInt() throws IOException {
    return Integer.parseInt(next());
  }

  static int[] nextIntArray(int len, int start) throws IOException {
    int[] a = new int[len];
    for (int i = start; i < len; i++)
      a[i] = nextInt();
    return a;
  }

  static long nextLong() throws IOException {
    return Long.parseLong(next());
  }

  static long[] nextLongArray(int len, int start) throws IOException {
    long[] a = new long[len];
    for (int i = start; i < len; i++)
      a[i] = nextLong();
    return a;
  }

  static String next() throws IOException {
    while (tok == null || !tok.hasMoreTokens()) {
      tok = new StringTokenizer(in.readLine());
    }
    return tok.nextToken();
  }

}
