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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

/**
 * Created by pallavi on 24/11/15.
 */
public class B602 {
    private static int n;
    private static int[] a;
    public static void main(String[] args) throws IOException {
        BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
        n = Integer.parseInt(reader.readLine());
        a = new int[n];
        String[] s = reader.readLine().split(" ");
        for (int i = 0; i < n; i++) {
            a[i] = Integer.parseInt(s[i]);
        }
        int p = 0;
        int m = a[0];
        int mi = 0, Mi = 0;
        int M = a[0];
        int c = 1;
        int cc = 1;
        for (int i = 1; i < n; i++) {
            if (m == M) {
                if (a[i] <= m) {
                    m = a[i];
                    mi = i;
                }
                if (a[i] >= M) {
                    M = a[i];
                    Mi = i;
                }
                cc += 1;
            } else {
                if (a[i] == m) {
                    m = a[i];
                    mi = i;
                    cc +=1;
                } else if (a[i] == M) {
                    M = a[i];
                    Mi = i;
                    cc +=1;
                }
                if (a[i] < m) {
                    cc = i-Mi;
                    M = m;
                    Mi = mi;
                    m = a[i];
                    mi = i;
                }
                if (a[i] > M) {
                    cc = i-mi;
                    m = M;
                    mi = Mi;
                    M = a[i];
                    Mi = i;
                }
            }
            c = Math.max(c, cc);

        }
        System.out.println(c);
    }
}
