#include <bits/stdc++.h>
using namespace std;
const int MAXN = 2e3 + 5;
int max_func(int a, int b, int c, int d, int e, int f) {
  return max(a, max(b, max(c, max(d, max(e, f)))));
}
int dp[MAXN][MAXN][3], rev_dp[MAXN][MAXN][3];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  int n;
  cin >> n;
  int arr[n];
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
  }
  for (int i = 0; i < n; i++) {
    for (int j = i; j < n; j++) {
      if (i == j) {
        if (arr[j] == 1)
          dp[i][j][0] = 1;
        else
          dp[i][j][2] = 1;
        continue;
      }
      if (arr[j] == 1) {
        dp[i][j][0] = dp[i][j - 1][0] + 1;
        dp[i][j][1] = dp[i][j - 1][1];
        dp[i][j][2] = dp[i][j - 1][2];
      } else {
        dp[i][j][0] = dp[i][j - 1][0];
        dp[i][j][1] = max(dp[i][j - 1][0], dp[i][j - 1][1]) + 1;
        dp[i][j][2] = dp[i][j - 1][2] + 1;
      }
    }
  }
  for (int i = n - 1; i >= 0; i--) {
    for (int j = i; j >= 0; j--) {
      if (i == j) {
        if (arr[j] == 1)
          rev_dp[i][j][0] = 1;
        else
          rev_dp[i][j][2] = 1;
        continue;
      }
      if (arr[j] == 1) {
        rev_dp[i][j][0] = rev_dp[i][j + 1][0] + 1;
        rev_dp[i][j][1] = rev_dp[i][j + 1][1];
        rev_dp[i][j][2] = rev_dp[i][j + 1][2];
      } else {
        rev_dp[i][j][0] = rev_dp[i][j + 1][0];
        rev_dp[i][j][1] = max(rev_dp[i][j + 1][0], rev_dp[i][j + 1][1]) + 1;
        rev_dp[i][j][2] = rev_dp[i][j + 1][2] + 1;
      }
    }
  }
  int ans = 0;
  for (int l = 0; l < n; l++) {
    for (int r = l; r < n; r++) {
      int a0, a1, a2;
      if (l == 0) {
        a0 = a1 = a2 = 0;
      } else {
        a0 = dp[0][l - 1][0], a1 = dp[0][l - 1][1], a2 = dp[0][l - 1][2];
      }
      int b0 = rev_dp[r][l][0], b1 = rev_dp[r][l][1], b2 = rev_dp[r][l][2];
      int c0, c1, c2;
      if (r == n - 1)
        c0 = c1 = c2 = 0;
      else {
        c0 = dp[r + 1][n - 1][0], c1 = dp[r + 1][n - 1][1],
        c2 = dp[r + 1][n - 1][2];
      }
      int d0 = a0 + b0 + c0;
      int d1 = a0 + b0 + c1;
      int d2 = a0 + b1 + c2;
      int d3 = a0 + b2 + c2;
      int d4 = a1 + b2 + c2;
      int d5 = a2 + b2 + c2;
      ans = max(ans, max_func(d0, d1, d2, d3, d4, d5));
    }
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Generate 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 MAXN = 2e3 + 5;
int max_func(int a, int b, int c, int d, int e, int f) {
  return max(a, max(b, max(c, max(d, max(e, f)))));
}
int dp[MAXN][MAXN][3], rev_dp[MAXN][MAXN][3];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  int n;
  cin >> n;
  int arr[n];
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
  }
  for (int i = 0; i < n; i++) {
    for (int j = i; j < n; j++) {
      if (i == j) {
        if (arr[j] == 1)
          dp[i][j][0] = 1;
        else
          dp[i][j][2] = 1;
        continue;
      }
      if (arr[j] == 1) {
        dp[i][j][0] = dp[i][j - 1][0] + 1;
        dp[i][j][1] = dp[i][j - 1][1];
        dp[i][j][2] = dp[i][j - 1][2];
      } else {
        dp[i][j][0] = dp[i][j - 1][0];
        dp[i][j][1] = max(dp[i][j - 1][0], dp[i][j - 1][1]) + 1;
        dp[i][j][2] = dp[i][j - 1][2] + 1;
      }
    }
  }
  for (int i = n - 1; i >= 0; i--) {
    for (int j = i; j >= 0; j--) {
      if (i == j) {
        if (arr[j] == 1)
          rev_dp[i][j][0] = 1;
        else
          rev_dp[i][j][2] = 1;
        continue;
      }
      if (arr[j] == 1) {
        rev_dp[i][j][0] = rev_dp[i][j + 1][0] + 1;
        rev_dp[i][j][1] = rev_dp[i][j + 1][1];
        rev_dp[i][j][2] = rev_dp[i][j + 1][2];
      } else {
        rev_dp[i][j][0] = rev_dp[i][j + 1][0];
        rev_dp[i][j][1] = max(rev_dp[i][j + 1][0], rev_dp[i][j + 1][1]) + 1;
        rev_dp[i][j][2] = rev_dp[i][j + 1][2] + 1;
      }
    }
  }
  int ans = 0;
  for (int l = 0; l < n; l++) {
    for (int r = l; r < n; r++) {
      int a0, a1, a2;
      if (l == 0) {
        a0 = a1 = a2 = 0;
      } else {
        a0 = dp[0][l - 1][0], a1 = dp[0][l - 1][1], a2 = dp[0][l - 1][2];
      }
      int b0 = rev_dp[r][l][0], b1 = rev_dp[r][l][1], b2 = rev_dp[r][l][2];
      int c0, c1, c2;
      if (r == n - 1)
        c0 = c1 = c2 = 0;
      else {
        c0 = dp[r + 1][n - 1][0], c1 = dp[r + 1][n - 1][1],
        c2 = dp[r + 1][n - 1][2];
      }
      int d0 = a0 + b0 + c0;
      int d1 = a0 + b0 + c1;
      int d2 = a0 + b1 + c2;
      int d3 = a0 + b2 + c2;
      int d4 = a1 + b2 + c2;
      int d5 = a2 + b2 + c2;
      ans = max(ans, max_func(d0, d1, d2, d3, d4, d5));
    }
  }
  cout << ans << endl;
  return 0;
}

```