Coderforces 933 A. A Twisty Movement

A. A Twisty Movement

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

A dragon symbolizes wisdom, power and wealth. On Lunar New Year's Day, people model a dragon with bamboo strips and clothes, raise them with rods, and hold the rods high and low to resemble a flying dragon.

A performer holding the rod low is represented by a 1, while one holding it high is represented by a 2. Thus, the line of performers can be represented by a sequence a1, a2, ..., an.

Little Tommy is among them. He would like to choose an interval [l, r] (1 ≤ l ≤ r ≤ n), then reverse al, al + 1, ..., ar so that the length of the longest non-decreasing subsequence of the new sequence is maximum.

A non-decreasing subsequence is a sequence of indices p1, p2, ..., pk, such that p1 < p2 < ... < pk and ap1 ≤ ap2 ≤ ... ≤ apk. The length of the subsequence is k.

Input

The first line contains an integer n (1 ≤ n ≤ 2000), denoting the length of the original sequence.

The second line contains n space-separated integers, describing the original sequence a1, a2, ..., an (1 ≤ ai ≤ 2, i = 1, 2, ..., n).

Output

Print a single integer, which means the maximum possible length of the longest non-decreasing subsequence of the new sequence.

Examples

input

Copy

4
1 2 1 2

output

Copy

4

input

Copy

10
1 1 2 2 2 1 1 2 2 1

output

Copy

9

Note

In the first example, after reversing [2, 3], the array will become [1, 1, 2, 2], where the length of the longest non-decreasing subsequence is 4.

In the second example, after reversing [3, 7], the array will become [1, 1, 1, 1, 2, 2, 2, 2, 2, 1], where the length of the longest non-decreasing subsequence is 9.

 

题意：给你一个由1和2组成的数列，你可以旋转一个字串，使得最长不下降子序列最长。

 

题解：现在题都这么神吗QAQ。这是A题好不好呀QAQ。我真是辣鸡鸭QAQ。

旋转前的序列一定是  11…1  + 22…2 + 11…1 + 22…2  这种形式的。

所以我们找个分割点 i 。1--i求个最长不下降，(i+1)--n求个最长不下降。加起来即可。

#include<bits/stdc++.h>
using namespace std;
int a[2010],f[2010][3],g[2010][3];
int main()
{
	int n,i,ans=0;
	scanf("%d",&n);
	for(i=1;i<=n;i++)scanf("%d",&a[i]);
	
	//f[i][j]为前i位(最大值为j)的最长不下降 
	for(i=1;i<=n;i++)
	{
		if(a[i]==1)
		{
			f[i][1]=f[i-1][1]+1;
			f[i][2]=f[i-1][2];
		}
		else
		{
			f[i][1]=f[i-1][1];
			f[i][2]=max(f[i-1][1],f[i-1][2])+1;
		}
	}
	
	//g[i][j]为后i位(最小值为j)的最长不下降 
	for(i=n;i>=1;i--)
	{
		if(a[i]==1)
		{
			g[i][1]=max(g[i+1][1],g[i+1][2])+1;
			g[i][2]=g[i+1][2];
		}
		else
		{
			g[i][1]=g[i+1][1];
			g[i][2]=g[i+1][2]+1;
		}
	}
	
	ans=0;
	for(i=1;i<=n;i++)
	{
		ans=max(ans,max(f[i][1],f[i][2])+max(g[i+1][1],g[i+1][2]));
	}
	printf("%d",ans);
	return 0;
}