cf462 A Twisty Movement

標準動態規劃
dp [i][j][0] 從i 到 j 的有多少1結尾的非遞增子序列 （ 21 或 1 ）
dp [i][j][1] 從i 到 j 的有多少2結尾的非遞增子序列 只有2
處理一下 多少個1 多少個 2 ，就可以get
 
C. A Twisty Movement

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, ..., a_n.

Little Tommy is among them. He would like to choose an interval [l, r] (1 ≤ l ≤ r ≤ n), then reverse a_l, a_l + 1, ..., a_r 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, ..., p_k, such that p₁ < p₂ < ... < p_k and a_p1 ≤ a_p2 ≤ ... ≤ a_pk. 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, ..., a_n (1 ≤ a_i ≤ 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

4
1 2 1 2

Output

4

Input

10
1 1 2 2 2 1 1 2 2 1

Output

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.

 

#include<iostream>
#include<algorithm>
using namespace std;
typedef long long ll;
#define f(i,l,r) for(int i=l;i<=r;i++)
#define g(i,l,r) for(int i=l;i>=r;i--)
const int maxn  = 2005 ; 

int a[maxn],b[maxn],c[maxn], n,ans= -2147483647;
int dp[maxn][maxn][2];
int main()
{
 
	cin>>n;
	f(i,1,n)cin>>a[i]; 
 	f(i,1,n) b[i]=b[i-1]+(a[i]==1);
	g(i,n,1) c[i]=c[i-1]+(a[i]==2) ;

	f(i,1,n)
	{
		f(j,1,n)
		{
			dp[i][j][0]=dp[i][j-1][0]+(a[i]==1);
			dp[i][j][1]=max(dp[i][j-1][1],d[i][j-1][0])+(a[i]==2);
			ans=max(ans,dp[i][j][0]+b[i-1]+c[j+1]);
			ans=max(ans,dp[i][j][1]+b[i-1]+c[j+1]); 
		}
	}

	cout<<ans<<endl;
 

	return 0;

}

    return 0;

}

未來的我一定會感謝現在正在成長的我