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;
struct debugger {
  template <typename T>
  debugger& operator,(const T& v) {
    cerr << v << " ";
    return *this;
  }
} debug;
int n;
int main() {
  do {
    std::ios::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
  } while (0);
  int n, tmp;
  int a[100005];
  cin >> n;
  for (__typeof(n) i = (0) - ((0) > (n)); i != (n) - ((0) > (n));
       i += 1 - 2 * ((0) > (n))) {
    cin >> tmp;
    a[i] = tmp;
  }
  int i = 1, j = 0, k = 0, max_d = 0;
  int valid[3] = {0, a[0], 0};
  for (; i < n; ++i) {
    if (a[i] == valid[1]) continue;
    if ((a[i] == (valid[1] + 1)) && !valid[0]) {
      valid[2] = a[i];
      continue;
    }
    if ((a[i] == (valid[1] - 1)) && !valid[2]) {
      valid[0] = a[i];
      continue;
    }
    max_d = max(max_d, i - j);
    j = i - 1;
    while (abs(a[i] - a[j]) <= 1) {
      --j;
    }
    ++j;
    valid[0] = (a[i] == a[j] - 1) ? a[i] : 0;
    valid[1] = a[j];
    valid[2] = (a[i] == a[j] + 1) ? a[i] : 0;
  }
  max_d = max(max_d, i - j);
  cout << max_d << endl;
  return 0;
}
