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.

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).

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

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：求最长不递减子序列可以选择一个区间逆转。

题解：求出1的前缀和，2的后缀和，以及区间[i,j]的最长不递增子序列。

f[i][j][0]表示区间i-j以1结尾的最长不递增子序列；

f[i][j][1]表示区间i-j以2结尾的最长不递增子序列，显然是区间i-j 2的个数；

所以转移方程为：

f[i][j][1] = f[i][j-1][1] + (a[j]==2);
   f[i][j][0] = max(f[i][j-1][0], f[i][j-1][1]) + (a[j]==1);（1<=i<=n,i<=j<=n）

代码：

 //#include"bits/stdc++.h"

 #include <sstream>

 #include <iomanip>

 #include"cstdio"

 #include"map"

 #include"set"

 #include"cmath"

 #include"queue"

 #include"vector"

 #include"string"

 #include"cstring"

 #include"time.h"

 #include"iostream"

 #include"stdlib.h"

 #include"algorithm"

 #define db double

 #define ll long long

 #define vec vector<ll>

 #define mt  vector<vec>

 #define ci(x) scanf("%d",&x)

 #define cd(x) scanf("%lf",&x)

 #define cl(x) scanf("%lld",&x)

 #define pi(x) printf("%d\n",x)

 #define pd(x) printf("%f\n",x)

 #define pl(x) printf("%lld\n",x)

 //#define rep(i, x, y) for(int i=x;i<=y;i++)

 #define rep(i,n) for(int i=0;i<n;i++)

 const int N   = 2e3 + ;

 const int mod = 1e9 + ;

 const int MOD = mod - ;

 const int inf = 0x3f3f3f3f;

 const db  PI  = acos(-1.0);

 const db  eps = 1e-;

 using namespace std;

 int a[N];

 int l[N],r[N];

 int f[N][N][];

 int main()

 {

     int n;

     ci(n);

     for(int i=;i<=n;i++) ci(a[i]),l[i]=l[i-]+(a[i]==);

     for(int i=n;i>=;i--) r[i]=r[i+]+(a[i]==);

     int ma=-;

     for(int i=;i<=n;i++){

         for(int j=i;j<=n;j++){

             f[i][j][]=f[i][j-][]+(a[j]==);

             f[i][j][]=max(f[i][j-][],f[i][j-][])+(a[j]==);

             ma=max(ma,f[i][j][]+l[i-]+r[j+]);

             ma=max(ma,f[i][j][]+l[i-]+r[j+]);

         }

     }

     pi(ma);

     return ;

 }