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;
char B[1 << 15], *S = B, *T = B;
inline int F() {
  register char ch;
  register int cnt = 0;
  register bool minus = 0;
  while (ch = (S == T && (T = (S = B) + fread(B, 1, 1 << 15, stdin), S == T)
                   ? EOF
                   : *S++),
         (ch < '0' || ch > '9') && ch != '-')
    ;
  ch == '-' ? minus = 1 : cnt = ch - '0';
  while (ch = (S == T && (T = (S = B) + fread(B, 1, 1 << 15, stdin), S == T)
                   ? EOF
                   : *S++),
         ch >= '0' && ch <= '9')
    cnt = cnt * 10 + ch - '0';
  return minus ? -cnt : cnt;
}
int a[100010], pos[100010];
int main() {
  register int n = F(), ans = 0;
  for (register int i = 1; i <= n; ++i) {
    register int x = F();
    pos[x] = i;
    if (pos[x - 1] < pos[x + 1])
      (ans < (i - ((pos[x - 1]) > (pos[x + 2]) ? (pos[x - 1]) : (pos[x + 2])))
           ? ans = (i -
                    ((pos[x - 1]) > (pos[x + 2]) ? (pos[x - 1]) : (pos[x + 2])))
           : 0);
    else
      (ans < (i - ((pos[x - 2]) > (pos[x + 1]) ? (pos[x - 2]) : (pos[x + 1])))
           ? ans = (i -
                    ((pos[x - 2]) > (pos[x + 1]) ? (pos[x - 2]) : (pos[x + 1])))
           : 0);
  }
  printf("%d\n", ans);
  return 0;
}
