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.*;
import java.util.*;

public class constant {

  public static void main(String[] args) {

      Scanner sc = new Scanner(System.in);

      int n = sc.nextInt();

      int[] a = new int[n];
      int[] diff = new int[n];

      for (int i = 0; i < n; i++) {
        a[i] = sc.nextInt();
        if (i > 0)
          diff[i] = a[i] - a[i - 1];
      }

      int l = 0;
      int r = 1;
      int last_diff = 0;
      int result = 2;
      int mark = 0;

      for (int i = 0; i < n; i++) {
        if (diff[i] != 0) {
          r = i;
          mark = i;
          last_diff = diff[i];
          break;
        }
        if (i == n - 1 && diff[i] == 0)
          result = i + 1;
      }

      if (r > result)
        result = r;

      while (r < n - 1) {
        r++;
        if (diff[r] != last_diff && diff[r] != 0) {
          last_diff = diff[r];
          mark = r;
          if (r - l + 1 > result)
            result = r - l + 1;
        }
        else if (diff[r] == last_diff) {
            if (r - l > result) {
              result = r - l;
            }
            l = mark;
            mark = r;
        }
        else if (diff[r] == 0) {
          if (r - l + 1 > result)
            result = r - l + 1;
        }

        // System.out.println(mark);
        //
        // System.out.print(l);
        // System.out.print(" ");
        // System.out.println(r);

      }

      System.out.println(result);

      sc.close();


  }

}
