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].
// package com.peterdang.hackerrank;

import java.io.IOException;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashSet;
import java.util.LinkedList;
import java.util.Scanner;

public class Hackerrank {

    public static void main(String[] args) throws IOException {
//        ArrayList<Integer> B = new ArrayList();

        int n;
        Scanner reader = new Scanner(System.in);

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


        int max = 0;
        int l, m, M;
        int indexm, indexM;
        boolean isStart = true;
        indexM = indexm = l = 0;
        m = M = A[l];
        if(n == 1)
            System.out.println(1);

        for (int r = 1; r < n; r++) {
            // 5 5 5 5 ...
            if (isStart) {
                // 5 5 5 5 7
                if (Math.abs(A[r] - A[l]) > 1) {
                    l = r;
                    indexM = indexm = l = 0;
                    m = M = A[l];
                } else { // 5 5 5 6
                    if (A[r] > A[l]) {
                        M = A[r];
                        indexM = r;
                        isStart = false;
                    } else if (A[r] < A[l]) {
                        m = A[r];
                        indexm = r;
                        isStart = false;
                    } else {
                        indexM = indexm = r;
                    }
                    max = Math.max(max, r - l + 1);
                }
            } else if (A[r] > M) {
                m = M;
                M = A[r];
                l = indexm + 1;
                indexm = indexM;
                indexM = r;
            } else if (A[r] < m) {
                M = m;
                m = A[r];
                l = indexM + 1;
                indexM = indexm;
                indexm = r;
            } else {
                if (A[r] == m) {
                    indexm = r;
                } else {
                    indexM = r;
                }
                max = Math.max(max, r - l + 1);
            }
        }

        System.out.println(max);

        reader.close();

    }
}
