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 ApproximatingAConstantRange_333 {

  public static void main(String[] args) throws IOException {
    BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
    int n = Integer.parseInt(reader.readLine());
    String[] numStrs = reader.readLine().split(" ");
    int[] nums = new int[n];
    for (int i = 0; i < n; i++) {
      nums[i] = Integer.parseInt(numStrs[i]);
    }
    int ans = 0;
    int left = 0, right = 0;
    int min = nums[left], max = nums[left];

    Map<Integer,Integer> countMap = new HashMap<>();
    countMap.put(nums[left], 1);

    while(right < n) {
      if(!validRange(min, max)) {
        countMap.put(nums[left], countMap.get(nums[left]) - 1);
        left++;
        if(countMap.get(min) == 0) {
          min = Integer.MAX_VALUE;
        }
        if(countMap.get(max) == 0) {
          max = Integer.MIN_VALUE;
        }

        min = Math.min(min, nums[left]);
        max = Math.max(max, nums[left]);

        if(countMap.getOrDefault(nums[left] - 1, 0) != 0) {
          min = Math.min(min, nums[left] - 1);
        }
        if(countMap.getOrDefault(nums[left] + 1, 0) != 0) {
          max = Math.max(max, nums[left] + 1);
        }
      }else {
        ans = Math.max(ans, right - left + 1);
        right++;
        if(right < n) {
          countMap.put(nums[right], countMap.getOrDefault(nums[right], 0) + 1);
          min = Math.min(min, nums[right]);
          max = Math.max(max, nums[right]);
        }
      }
    }
    System.out.println(ans);
  }

  private static boolean validRange(int min, int max) {
    return max - min <= 1;
  }

}
