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;
void task();
int main() {
  task();
  return 0;
}
const int N = (int)1e5 + 10;
const long long int MODULO = (long long int)1e9 + 7;
const int INF = (int)1e9 + 10;
const int BUF_SIZE = (int)1e4 + 10;
int a[N];
void task() {
  ios_base::sync_with_stdio(0);
  int n;
  cin >> n;
  for (int i = 0; i < n; ++i) cin >> a[i];
  int pred = -1, now = a[1], ind = 0, ans = 0, tans = 1;
  int i = 1;
  for (; a[i] == a[i - 1]; ++i, ++tans)
    ;
  if (i == n) {
    cout << tans;
    return;
  }
  (void(1));
  ;
  pred = a[i - 1];
  now = a[i];
  ind = i;
  ++i, ++tans;
  for (; i < n; ++i) {
    (void(1));
    ;
    if (a[i] != a[i - 1]) {
      if (a[i] != pred) {
        ans = max(ans, tans);
        tans = i - ind;
      }
      pred = now;
      now = a[i];
      ind = i;
    }
    ++tans;
  }
  ans = max(ans, tans);
  cout << ans;
}
