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;

public class B {
    public static void main(String[] args) throws Exception {
        solution(new Scanner(System.in));
    }

    public static void solution(Scanner sc) {
        int n = sc.nextInt();
        int res = 0;
        int n1 = -1, n2 = -1, i1 = -1, i11 = -1, i2 = -1, i22 = -1;
        for (int i = 0; i < n; i++) {
            int ni = sc.nextInt();
            if (n1 == -1) {
                n1 = ni;
                i1 = i;
                i11 = i;
//                System.out.printf("Debug %d first n1\n", 0);
//                System.out.printf("n1 = %d [%d, %d] n2 = %d [%d, %d] \n", n1, i1,i11,n2,i2,i22);
                res = Math.max(res, i - i1 + 1);
            }
            else {
                if (ni == n1) {
                    i11 = i;
//                    System.out.printf("Debug %d add n1\n", 1);
//                    System.out.printf("n1 = %d [%d, %d] n2 = %d [%d, %d] \n", n1, i1,i11,n2,i2,i22);
                    res = Math.max(res, i - i1 + 1);
                }
                else {
                    if (n2 == -1) {
                        n2 = ni;
                        i2 = i;
                        i22 = i;
//                        System.out.printf("Debug %d first n2\n", 2);
//                        System.out.printf("n1 = %d [%d, %d] n2 = %d [%d, %d] \n", n1, i1,i11,n2,i2,i22);
                        res = Math.max(res, i - i1 + 1);
                    }
                    else {
                        if (ni == n2) {
                            i22 = i;
//                            System.out.printf("Debug %d add n2\n", 3);
//                            System.out.printf("n1 = %d [%d, %d] n2 = %d [%d, %d] \n", n1, i1,i11,n2,i2,i22);
                            res = Math.max(res, i - i1 + 1);
                        }
                        else {
                            if (Math.abs(ni - n2) <= 1 && i11 + 1 < i) {
                                n1 = n2;
                                i1 = i11 + 1;
                                i11 = i22;
                                n2 = ni;
                                i2 = i;
                                i22 = i;
//                                System.out.printf("Debug %d new n1, n2\n", 4);
//                                System.out.printf("n1 = %d [%d, %d] n2 = %d [%d, %d] \n", n1, i1,i11,n2,i2,i22);
                                res = Math.max(res, i - i1 + 1);
                            }
                            else if (Math.abs(ni - n1) <= 1 && i22 + 1 < i) {
                                i1 = i22 + 1;
                                i11 = Math.max(i22 + 1, i11);
                                n2 = ni;
                                i2 = i;
                                i22 = i;
//                                System.out.printf("Debug %d new n2\n", 5);
//                                System.out.printf("n1 = %d [%d, %d] n2 = %d [%d, %d] \n", n1, i1,i11,n2,i2,i22);
                                res = Math.max(res, i - i1 + 1);
                            }
                            else {
                                n1 = ni;
                                i1 = i;
                                i11 = i;
                                n2 = -1;
                                i2 = -1;
                                i22 = -1;
//                                System.out.printf("Debug %d new n1\n", 6);
//                                System.out.printf("n1 = %d [%d, %d] n2 = %d [%d, %d] \n", n1, i1,i11,n2,i2,i22);
                                res = Math.max(res, i - i1 + 1);
                            }
                        }
                    }
                }
            }
//            System.out.printf("n[%d] = %d , res = %d \n\n", i, ni, res);
        }
        sc.close();
        System.out.println(res);
    }
}
