#include <bits/stdc++.h>
using namespace std;
const int INF = 1e9 + 7;
const int MAXN = 1e6 + 7;
const double EPS = 1e-8;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  vector<int> a(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    a[i]--;
  }
  vector<vector<vector<int> > > dp(
      n + 1, vector<vector<int> >(n + 1, vector<int>(2, 0)));
  vector<vector<int> > cnt1(n + 1, vector<int>(n + 1, 0));
  vector<vector<int> > cnt2(n + 1, vector<int>(n + 1, 0));
  for (int i = 0; i < n; i++) {
    int sum1 = 0, sum2 = 0;
    for (int j = i; j < n; j++) {
      if (a[j]) {
        sum2++;
      } else {
        sum1++;
      }
      cnt1[i][j] = sum1;
      cnt2[i][j] = sum2;
    }
  }
  for (int i = n - 1; i >= 0; i--) {
    int sum1 = 0, sum2 = 0;
    for (int j = i; j >= 0; j--) {
      if (!a[j]) {
        sum1++;
      } else {
        sum2++;
      }
      sum2 = max(sum2, sum1);
      dp[j][i][0] = sum1;
      dp[j][i][1] = sum2;
    }
  }
  int sum1 = 0, sum2 = 0;
  for (int i = 0; i < n; i++) {
    if (a[i]) {
      sum2++;
    } else {
      sum1++;
    }
    sum2 = max(sum2, sum1);
  }
  int maxk = sum2;
  for (int l = 0; l < n; l++) {
    int sum1 = (l ? cnt1[0][l - 1] : 0);
    for (int r = l; r < n; r++) {
      maxk = max(maxk, sum1 + dp[l][r][1] + cnt2[r + 1][n - 1]);
    }
  }
  cout << maxk;
}

### Prompt

Please formulate a Cpp solution to the following problem:

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

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int INF = 1e9 + 7;
const int MAXN = 1e6 + 7;
const double EPS = 1e-8;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  vector<int> a(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    a[i]--;
  }
  vector<vector<vector<int> > > dp(
      n + 1, vector<vector<int> >(n + 1, vector<int>(2, 0)));
  vector<vector<int> > cnt1(n + 1, vector<int>(n + 1, 0));
  vector<vector<int> > cnt2(n + 1, vector<int>(n + 1, 0));
  for (int i = 0; i < n; i++) {
    int sum1 = 0, sum2 = 0;
    for (int j = i; j < n; j++) {
      if (a[j]) {
        sum2++;
      } else {
        sum1++;
      }
      cnt1[i][j] = sum1;
      cnt2[i][j] = sum2;
    }
  }
  for (int i = n - 1; i >= 0; i--) {
    int sum1 = 0, sum2 = 0;
    for (int j = i; j >= 0; j--) {
      if (!a[j]) {
        sum1++;
      } else {
        sum2++;
      }
      sum2 = max(sum2, sum1);
      dp[j][i][0] = sum1;
      dp[j][i][1] = sum2;
    }
  }
  int sum1 = 0, sum2 = 0;
  for (int i = 0; i < n; i++) {
    if (a[i]) {
      sum2++;
    } else {
      sum1++;
    }
    sum2 = max(sum2, sum1);
  }
  int maxk = sum2;
  for (int l = 0; l < n; l++) {
    int sum1 = (l ? cnt1[0][l - 1] : 0);
    for (int r = l; r < n; r++) {
      maxk = max(maxk, sum1 + dp[l][r][1] + cnt2[r + 1][n - 1]);
    }
  }
  cout << maxk;
}

```