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, k;
int a, la, fa, b, lb, fb, last;
int ma;
int main() {
  int M = 1000007;
  ma = 0;
  cin >> n;
  a = M;
  b = M;
  la = M;
  lb = M;
  fa = M;
  fb = M;
  for (int i = 0; i < n; i++) {
    cin >> k;
    ma = max(ma, i - min(fa, fb));
    if (k == a) {
      lb = M;
      fa = min(fa, i);
      la = min(la, i);
    } else if (k == b) {
      la = M;
      fb = min(fb, i);
      lb = min(lb, i);
    } else if (k == a + 1 || k == a - 1) {
      b = k;
      fb = i;
      lb = i;
      fa = la;
      la = M;
    } else if (k == b + 1 || k == b - 1) {
      a = k;
      fa = i;
      la = i;
      fb = lb;
      lb = M;
    } else {
      a = k;
      fa = i;
      la = i;
      b = M;
      fb = M;
      lb = M;
    }
    last = k;
  }
  ma = max(ma, n - min(fa, fb));
  cout << ma;
}
