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>
using namespace std;
int n, a[100014], dem[100014], val1, val2, ans, l;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = (1); i <= (n); i++) cin >> a[i];
  dem[a[1]]++;
  dem[a[2]]++;
  ans = 2;
  val1 = a[1];
  val2 = a[2];
  l = 1;
  for (int i = (3); i <= (n); i++) {
    if (a[i] != val1 && a[i] != val2 && val1 != val2) {
      do {
        dem[a[l]]--;
        if (dem[a[l]] == 0) {
          if (a[l] == val1)
            val1 = a[i];
          else
            val2 = a[i];
          l++;
          break;
        }
        l++;
      } while (1);
    }
    if (val1 == val2 && val1 != a[i]) val1 = a[i];
    ans = max(ans, i - l + 1);
    dem[a[i]]++;
  }
  cout << ans;
  return 0;
}
