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 double pi = acos(-1.0);
const double eps = 1e-11;
int main() {
  int n;
  cin >> n;
  vector<int> v(n);
  for (int i = 0; i < n; ++i) {
    cin >> v[i];
  }
  int mx = 0;
  deque<pair<int, int> > Q1;
  deque<pair<int, int> > Q2;
  for (int i = 0; i < n; i++) {
    if (Q1.empty() || Q1.front().second == v[i]) {
      Q1.push_front(make_pair(i, v[i]));
    } else if (Q2.empty() || Q2.front().second == v[i]) {
      Q2.push_front(make_pair(i, v[i]));
    } else if (v[i] != Q1.front().second && v[i] != Q2.front().second) {
      if (abs(Q1.front().second - v[i]) <= 1) {
        int mx = Q2.front().first;
        while (!Q2.empty()) {
          Q2.pop_back();
        }
        Q2.push_front(make_pair(i, v[i]));
        while (!Q1.empty() && Q1.back().first <= mx) Q1.pop_back();
      } else {
        int mx = Q1.front().first;
        while (!Q1.empty()) {
          Q1.pop_back();
        }
        Q1.push_front(make_pair(i, v[i]));
        while (!Q2.empty() && Q2.back().first <= mx) Q2.pop_back();
      }
    }
    mx = max(mx, int((Q1).size()) + int((Q2).size()));
  }
  cout << mx << endl;
  return 0;
}
