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.util.Scanner;
public class approximatingRange {
  public static void main(String[] args){
    Scanner scan = new Scanner(System.in);
    int n, a, min, max, minpos, maxpos, len, pos;
    n = scan.nextInt();
   // int[] a = new int[n];
    pos = minpos = maxpos = len = 0;
    min = max = scan.nextInt();
    for(int i = 1;i < n;i++){
      a = scan.nextInt();
      if(Math.abs(a - min) <= 1 && Math.abs(a - max) <= 1){
        if(a == min && a == max){
        min = max = a;
        minpos = maxpos = i;
        }
        else if(a>=max){
        max = a;
        maxpos = i;
   
        }
        else if(a<=min){
        min = a;
        minpos = i;
          }
        }
      else if(a == max + 1){
        len = Math.max(len, i - pos);
        max = a;
        if(maxpos == i-1){ 
          min = a - 1;
          pos = minpos + 1;
          minpos = i-1;
          maxpos = i;
        }
        else {
          min = a;
          pos = minpos = maxpos = i;
        }
      }
      else if(a == min-1){
        len = Math.max(len, i - pos);
        min = a;
        if(minpos == i-1){
        max = a + 1;
        pos = maxpos + 1;
        maxpos = i - 1;
        minpos = i;
        }
        else {
          max = a;
          pos = minpos = maxpos = i;
        }
         }
      else {
      len = Math.max(len, i - pos);      
      maxpos = minpos = i;
      pos = i;
      min = max = a;
      }
    }
    len = Math.max(len, n - pos);
       System.out.println(len);
  }
}