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;
import java.util.TreeSet;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int a[] = new int[n];
        for(int i = 0; i < n; ++i)
            a[i] = sc.nextInt();

        TreeSet<Element> s = new TreeSet<Element>((lhs, rhs) -> {
            if (lhs.val < rhs.val) return -1;
            if (lhs.val > rhs.val) return 1;
            if (lhs.index < rhs.index) return -1;
            if (lhs.index > rhs.index) return 1;
            return 0;
        });

        int l = 0, ans = 0;

        for(int r = 0; r < n; ++r) {
            s.add(new Element(a[r], r));
            while (s.last().val - s.first().val > 1) {
                s.remove(new Element(a[l], l));
                ++l;
            }
            ans = Math.max(ans, r-l+1);
        }

        System.out.println(ans);
    }

    public static class Element {
        int val, index;

        public Element(int val, int index) {
            this.val = val;
            this.index = index;
        }
    }
}




