Codeforces Round #462 (Div. 2) C DP

C. 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

4

input

Copy

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.

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
",x)
#define pd(x) printf("%f
",x)
#define pl(x) printf("%lld
",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 + 5;
const int mod = 1e9 + 7;
const int MOD = mod - 1;
const int inf = 0x3f3f3f3f;
const db  PI  = acos(-1.0);
const db  eps = 1e-10;
using namespace std;
int a[N];
int l[N],r[N];
int f[N][N][2];
int main()
{
    int n;
    ci(n);
    for(int i=1;i<=n;i++) ci(a[i]),l[i]=l[i-1]+(a[i]==1);
    for(int i=n;i>=0;i--) r[i]=r[i+1]+(a[i]==2);
    int ma=-1;
    for(int i=1;i<=n;i++){
        for(int j=i;j<=n;j++){
            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);
            ma=max(ma,f[i][j][1]+l[i-1]+r[j+1]);
            ma=max(ma,f[i][j][0]+l[i-1]+r[j+1]);
        }
    }
    pi(ma);
    return 0;
}