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.*;
import java.lang.Math;

public class B {

    static int[] arr;

    static int[] find_min(int i, int j) {
        int min_ = arr[i];
        int imin = i;
        for (int z=i+1; z<= j; z++) {
            if (min_ >= arr[z]) {
                min_ = arr[z];
                imin = z;
            }
        }
        int[] kk = new int[2];
        kk[0] = imin;
        kk[1] = min_;
        return kk;
    }
    
    static int[] find_max(int i, int j) {
        int max_ = arr[i];
        int imax = i;
        for (int z=i+1; z <= j; z++) {
            if (max_ <= arr[z]) {
                max_ = arr[z];
                imax = z;
            }
        }
        int[] kk = new int[2];
        kk[0] = imax;
        kk[1] = max_;
        return kk;
    }

    static int sol() {
        int n = arr.length;
        int a = 0;
        int b = 0;
        int maxl = 1;
        int max_ = arr[0];
        int imax = 0;
        int min_ = arr[0];
        int imin = 0;
        while (b < n) {
            int e = arr[b];
            if (e < min_) {
                min_ = e;
                imin = b;
            }
            if (e > max_) {
                max_ = e;
                imax = b;
            }

            while (max_ - min_ > 1) {
                if (imin < imax) {
                    int[] kk = find_min(imin + 1, imax);
                    a = imin + 1;
                    imin = kk[0];
                    min_ = kk[1];
                } else {
                    int[] kk = find_max(imax + 1, imin);
                    a = imax + 1;
                    imax = kk[0];
                    max_ = kk[1];
                }
            }
            if (b - a + 1 > maxl)
                maxl = b - a + 1;
            b++;
        }
        return maxl;
    }

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        arr = new int[n];
        for (int i=0; i<n; i++) {
            int e = sc.nextInt();
            arr[i] = e;
        }
        System.out.println(sol());

    }
}
