Codeforces Round #333 (Div. 2) B. Approximating a Constant Range （dp）

B. Approximating a Constant Range

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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

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].

题解:基础dp，尺取法也可以。

代码：

#pragma comment(linker, "/STACK:102400000,102400000")//#include<bits/stdc++.h>#include<stdio.h>#include<string.h>#include<algorithm>#include<iostream>#include<cstring>#include<vector>#include<map>#include<cmath>#include<queue>#include<set>#include<stack>#include <utility>using namespace std;typedef long long ll;typedef unsigned long long ull;#define mst(a) memset(a, 0, sizeof(a))#define M_P(x,y) make_pair(x,y)  #define rep(i,j,k) for (int i = j; i <= k; i++)  #define per(i,j,k) for (int i = j; i >= k; i--)  #define lson x << 1, l, mid  #define rson x << 1 | 1, mid + 1, r  const int lowbit(int x) { return x&-x; }  const double eps = 1e-8;  const int INF = 1e9+7; const ll inf =(1LL<<62) ;const int MOD = 1e9 + 7;  const ll mod = (1LL<<32);const int N = 3e5+7; const int M=100010;const int maxn=2e3+7; template <class T1, class T2>inline void getmax(T1 &a, T2 b) {if (b>a)a = b;}  template <class T1, class T2>inline void getmin(T1 &a, T2 b) {if (b<a)a = b;}int read(){int v = 0, f = 1;char c =getchar();while( c < 48 || 57 < c ){if(c=='-') f = -1;c = getchar();}while(48 <= c && c <= 57) v = v*10+c-48, c = getchar();return v*f;}int dp[1000010]; int main(){int ans=1,n,a,l=0;cin>>n;for(int i=1;i<=n;++i){scanf("%d",&a);++a;l=max(l,dp[a-2]);l=max(l,dp[a+2]);dp[a]=i;ans=max(ans,i-l);}cout<<ans;return 0;}