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;
vector<int> nums[100001];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    int inp;
    scanf("%d", &inp);
    inp--;
    nums[inp].push_back(i);
  }
  int mx_len = 0;
  for (int i = 0; i < 100000; i++) {
    if (nums[i].size()) {
      vector<int> vec = nums[i];
      int last = 0, sz = vec.size();
      for (int j = 1; j < sz; j++) {
        if (vec[j] - vec[j - 1] != 1) {
          int temp_len = j - last;
          last = j;
          mx_len = max(mx_len, temp_len);
        }
      }
      if (last != sz - 1) mx_len = max(mx_len, sz - last);
    }
    if (nums[i].size() && nums[i + 1].size()) {
      vector<int> vec = nums[i];
      vec.insert(vec.end(), nums[i + 1].begin(), nums[i + 1].end());
      sort(vec.begin(), vec.end());
      int last = 0, sz = vec.size();
      for (int j = 1; j < sz; j++) {
        if (vec[j] - vec[j - 1] != 1) {
          int temp_len = j - last;
          last = j;
          mx_len = max(mx_len, temp_len);
        }
      }
      if (last != sz - 1) mx_len = max(mx_len, sz - last);
    }
  }
  printf("%d", mx_len);
  return 0;
}
