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.*;

public class Constant {
    public static void main(String[] args) {
        Scanner scan = new Scanner(System.in);

        int n = scan.nextInt();
        int[] data = new int[n];
        
        for(int i = 0; i < n; i++) {
            data[i] = scan.nextInt();
        }

        int maxVal = data[0], minVal = data[0];
        int minIndex = 0, maxIndex = 0;
        int currentLength = 1;
        int maxLength = 1;
        for(int i = 1; i < n; i++) {
            if(maxVal <= data[i]) {
                maxVal = data[i];
                maxIndex = i;
            }
            if(minVal >= data[i]) {
                minVal = data[i];
                minIndex = i;
            }

            if(maxVal - minVal <= 1) {
                currentLength++;
                if (i == n - 1) {
                    maxLength = (int)Math.max(maxLength, currentLength);    
                }
            }
            else {
                maxLength = (int)Math.max(maxLength, currentLength);
                if(minIndex < maxIndex) {
                    currentLength = maxIndex - minIndex;
                    minVal = data[minIndex+1];
                    minIndex += 1;
                    i = maxIndex;
                }
                else {
                    currentLength = minIndex - maxIndex;
                    maxVal = data[maxIndex + 1];
                    maxIndex += 1;
                    i = minIndex;
                }
            }
        }
        System.out.println(maxLength);
    }
}