CodeForces 602B Approximating a Constant Range（RMQ）

题意：给你n个数，要求你找到最长的区间，使得这个区间的最大值减去最小值之差的绝对值小于等于1

思路：枚举每一个数，以这个数为这个区间的最小值，能够往左边延伸多少，往右边延伸多少，再枚举每一个数，以这个数为区间的最大值，能够往左边延伸多少，往右边延伸多少就好了

#include<bits\stdc++.h>using namespace std;const int maxn = 100000+100;int n;int L[maxn],R[maxn];int dmax[maxn][20];int dmin[maxn][20];int d[maxn];void initmax(int n,int d[]){for (int i = 1;i<=n;i++)dmax[i][0]=d[i];for (int j = 1;(1<<j)<=n;j++)for (int i = 1;i+(1<<j)-1<=n;i++)dmax[i][j]=max(dmax[i][j-1],dmax[i+(1<<(j-1))][j-1]);}int getmax(int L,int R){int k = 0;while ((1<<(k+1)) <= R-L+1)k++;return max(dmax[L][k],dmax[R-(1<<k)+1][k]);}void initmin(int n,int d[]){for (int i = 1;i<=n;i++)dmin[i][0]=d[i];for (int j = 1;(1<<j)<=n;j++)for (int i = 1;i+(1<<j)-1<=n;i++)dmin[i][j]=min(dmin[i][j-1],dmin[i+(1<<(j-1))][j-1]);}int getmin(int L,int R){int k = 0;while ((1<<(k+1)) <= R-L+1)k++;return min(dmin[L][k],dmin[R-(1<<k)+1][k]);}int main(){    int q;int ans = 0;scanf("%d",&n);for(int i = 1;i<=n;i++)scanf("%d",&d[i]);    initmax(n,d);initmin(n,d);    for(int i = 1;i<=n;i++){int l = 1,r=i;while(l<=r){int m = (l+r)>>1;if (abs(d[i]-getmin(m,i))<=0 && abs(d[i]-getmax(m,i))<=1)r=m-1;elsel=m+1;}L[i]=l;        l=i,r=n;while(l<=r){int m = (l+r)>>1;if (abs(d[i]-getmin(i,m))<=0&&abs(d[i]-getmax(i,m))<=1)l=m+1;elser=m-1;}R[i]=l-1;}for(int i = 1;i<=n;i++)ans = max(ans,R[i]-L[i]+1);printf("%d\n",ans);}

Description

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 Input

Input

51 2 3 3 2

Output

4

Input

115 4 5 5 6 7 8 8 8 7 6

Output

5

Hint

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].