B. Approximating a Constant Range

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, ..., a_n. There aren‘t any big jumps between consecutive data points — for each 1 ≤ i < n, it‘s guaranteed that |a_i + 1 - a_i| ≤ 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 a_i 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, ..., a_n (1 ≤ a_i ≤ 100 000).

Output

Print a single number — the maximum length of an almost constant range of the given sequence.

Sample test(s)

Input

51 2 3 3 2

Output

4

Input

115 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].

 1 #include<cstdio>
 2 #include<vector>
 3 #include<cmath>
 4 #include<queue>
 5 #include<map>
 6 #include<cstring>
 7 #include<algorithm>
 8 using namespace std;
 9 typedef long long ll;
10 typedef unsigned long long ull;
11 const int maxn=100005;
12 int main()
13 {
14     int n;
15     int a[maxn],s[maxn];
16     scanf("%d",&n);
17     for(int i=1;i<=n;i++)
18         scanf("%d",&a[i]);
19     int ans=1;
20     for(int i=1,l=1,r=0;i<=n;i++)
21     {
22         while(l<=r&&a[i]<a[s[r]])r--;
23         s[++r]=i;
24         while(a[s[r]]>a[s[l]]+1)l++;
25         ans=max(ans,s[r]-s[l]+1);
26     }
27     for(int i=1,l=1,r=0;i<=n;i++)
28     {
29         while(l<=r&&a[i]>a[s[r]])r--;
30         s[++r]=i;
31         while(a[s[l]]>a[s[r]]+1)l++;
32         ans=max(ans,s[r]-s[l]+1);
33     }
34     printf("%d\n",ans);
35     return 0;
36 }