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;
const long long int INF = 1e18;
const long long int mod = 1e9 + 7;
int main() {
  int n;
  cin >> n;
  vector<int> a(n);
  vector<int> c(n, 0);
  for (int i = 0; i <= n - 1; i++) cin >> a[i];
  pair<int, int> mi;
  pair<int, int> ma;
  mi.first = a[0];
  mi.second = 0;
  ma.first = a[0] + 1;
  ma.second = -1;
  int ans = 1;
  int count = 1;
  for (int i = 1; i <= n - 1; i++) {
    if (a[i] == mi.first) {
      mi.second = i;
      count++;
    } else if (a[i] == ma.first) {
      ma.second = i;
      count++;
    } else if (a[i] == mi.first - 1) {
      int temp1 = mi.first, temp2 = mi.second;
      int temp3 = ma.first, temp4 = ma.second;
      mi.first = a[i], mi.second = i;
      ma.first = temp1, ma.second = temp2;
      count = i - temp4;
    } else if (a[i] < mi.first - 1) {
      mi.first = a[i], mi.second = i;
      ma.first = mi.first + 1, ma.second = i - 1;
      count = 1;
    } else if (a[i] == ma.first + 1) {
      int temp1 = mi.first, temp2 = mi.second;
      int temp3 = ma.first, temp4 = ma.second;
      ma.first = a[i], ma.second = i;
      mi.first = temp3, mi.second = temp4;
      count = i - temp2;
    } else if (a[i] > ma.first + 1) {
      mi.first = a[i], mi.second = i;
      ma.first = mi.first + 1, ma.second = i - 1;
      count = 1;
    }
    ans = max(count, ans);
  }
  cout << ans << endl;
  return 0;
}
