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 main() {
  int n;
  cin >> n;
  vector<int> mas(n);
  for (int i = 0; i < n; i++) {
    cin >> mas[i];
  }
  int l = 0, r = 0;
  int max = 1;
  int small, big;
  while (r < n && mas[r] == mas[0]) {
    r++;
  }
  if (r == n) {
    cout << n;
    return 0;
  }
  int lastS, lastB;
  if (mas[l] < mas[r]) {
    small = mas[l];
    big = mas[r];
    lastS = r - 1;
    lastB = r;
  } else {
    small = mas[r];
    big = mas[l];
    lastS = r;
    lastB = r - 1;
  }
  r++;
  while (r < n && small <= mas[r] && mas[r] <= big) {
    if (mas[r] == small) {
      lastS = r;
    } else {
      lastB = r;
    }
    r++;
  }
  if (r == n) {
    cout << n;
    return 0;
  }
  while (r < n) {
    if (r - l > max) {
      max = r - l;
    }
    if (mas[r] < small) {
      l = lastB + 1;
      big = small;
      small = mas[r];
      if (l > lastS) {
        lastB = l;
      } else {
        lastB = lastS;
      }
      lastS = r;
    } else if (mas[r] > big) {
      l = lastS + 1;
      small = big;
      big = mas[r];
      if (l > lastB) {
        lastS = l;
      } else {
        lastS = lastB;
      }
      lastB = r;
    } else {
      if (mas[r] == small) {
        lastS = r;
      } else {
        lastB = r;
      }
    }
    r++;
    while (r < n && small <= mas[r] && mas[r] <= big) {
      if (mas[r] == small) {
        lastS = r;
      } else {
        lastB = r;
      }
      r++;
    }
  }
  if (r - l > max) {
    cout << r - l;
  } else {
    cout << max;
  }
  return 0;
}
