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].
#include <bits/stdc++.h>
#pragma warning(disable : 4996)
using namespace std;
int n, a[100010];
int b[100010];
int main() {
  cin >> n;
  int m = 0;
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    b[i] = 1;
  }
  int max = 0;
  for (int i = 0; i < n; i++) {
    bool judge1 = true, judge2 = true;
    int j = i + 1, min = a[i], count = 0;
    while ((a[j] - min <= 1 && a[j] - min >= 0) ||
           (abs(a[j] - min) <= 1 && judge2) && j < n) {
      if (a[j] != a[j - 1]) {
        judge2 = false;
        judge1 = false;
        min = a[j] > min ? min : a[j];
      } else if (judge1) {
        count++;
      }
      j++;
      b[m]++;
    }
    if (b[m] > max) {
      max = b[m];
    }
    m++;
    i += count;
    if (j == n) {
      break;
    }
  }
  printf("%d", max);
  return 0;
}
