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.Scanner;
import java.util.ArrayList;
import java.util.Arrays;
import java.lang.Math;

public class MyCode {
	public static void main (String[] args) {
      Scanner sc = new Scanner(System.in);
      int n = sc.nextInt();

      ArrayList<Integer> list = new ArrayList<>();

      for (int i = 0; i < n; i++){
        list.add(sc.nextInt());
      }
      
      int[] a = new int[100001];
      int l = 0;
      int r = 0;
      int count = 0;
      int ans = 0;
      for (int i = r; i < n; i++){
        int s = list.get(i);
        if(count < 2){
          if(a[s] == 0){
            a[s] += 1;
            r = i;
            count += 1;
          } else {
            a[s] += 1;
            r = i;
          }
        } else if (count == 2) {
          if(a[s] == 0){
            ans = Math.max(ans, r - l + 1);
            while (true){
              a[list.get(l)] -= 1;
              if(a[list.get(l)] == 0){
                l += 1;
                break;
              } else {
                l += 1;
              }
            }
            a[list.get(i)] += 1;
            r = i;
          } else {
            a[list.get(i)] += 1;
            r = i;
          }
        }
      }
      ans = Math.max(ans, r - l + 1);
      
      System.out.print(ans);
  }
}