#include <bits/stdc++.h>
using namespace std;
static inline void canhazfast() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
}
template <typename T>
T gcd(T a, T b) {
  return b == 0 ? a : gcd(b, a % b);
}
int main() {
  canhazfast();
  static int a[2000 + 1];
  static int f[2000 + 2][2], b[2000 + 2][2];
  int n;
  int ans = 0;
  cin >> n;
  for (int i = 1; i <= n; ++i) cin >> a[i];
  for (int i = 1; i <= n; ++i) {
    if (a[i] == 1) {
      f[i][0] = f[i - 1][0] + 1;
      f[i][1] = f[i - 1][1];
    } else {
      f[i][0] = f[i - 1][0];
      f[i][1] = max(f[i - 1][0], f[i - 1][1]) + 1;
    }
  }
  for (int i = n; i > 0; --i) {
    if (a[i] == 1) {
      b[i][0] = max(b[i + 1][0], b[i + 1][1]) + 1;
      b[i][1] = b[i + 1][1];
    } else {
      b[i][0] = b[i + 1][0];
      b[i][1] = b[i + 1][1] + 1;
    }
  }
  for (int r = n; r > 0; --r) {
    int c[2] = {};
    int prev[2] = {};
    int next[2];
    for (int l = r; l > 0; --l) {
      ++c[a[l] - 1];
      if (a[l] == 1) {
        next[0] = prev[0] + 1;
        next[1] = prev[1];
      } else {
        next[0] = prev[0];
        next[1] = max(prev[0], prev[1]) + 1;
      }
      ans = max(ans, f[l - 1][0] + c[0] + max(b[r + 1][0], b[r + 1][1]));
      ans = max(ans, f[l - 1][0] + next[1] + b[r + 1][1]);
      ans = max(ans, max(f[l - 1][0], f[l - 1][1]) + c[1] + b[r + 1][1]);
      prev[0] = next[0];
      prev[1] = next[1];
    }
  }
  cout << ans << '\n';
  return 0;
}

### Prompt

Create a solution in cpp for 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;
static inline void canhazfast() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
}
template <typename T>
T gcd(T a, T b) {
  return b == 0 ? a : gcd(b, a % b);
}
int main() {
  canhazfast();
  static int a[2000 + 1];
  static int f[2000 + 2][2], b[2000 + 2][2];
  int n;
  int ans = 0;
  cin >> n;
  for (int i = 1; i <= n; ++i) cin >> a[i];
  for (int i = 1; i <= n; ++i) {
    if (a[i] == 1) {
      f[i][0] = f[i - 1][0] + 1;
      f[i][1] = f[i - 1][1];
    } else {
      f[i][0] = f[i - 1][0];
      f[i][1] = max(f[i - 1][0], f[i - 1][1]) + 1;
    }
  }
  for (int i = n; i > 0; --i) {
    if (a[i] == 1) {
      b[i][0] = max(b[i + 1][0], b[i + 1][1]) + 1;
      b[i][1] = b[i + 1][1];
    } else {
      b[i][0] = b[i + 1][0];
      b[i][1] = b[i + 1][1] + 1;
    }
  }
  for (int r = n; r > 0; --r) {
    int c[2] = {};
    int prev[2] = {};
    int next[2];
    for (int l = r; l > 0; --l) {
      ++c[a[l] - 1];
      if (a[l] == 1) {
        next[0] = prev[0] + 1;
        next[1] = prev[1];
      } else {
        next[0] = prev[0];
        next[1] = max(prev[0], prev[1]) + 1;
      }
      ans = max(ans, f[l - 1][0] + c[0] + max(b[r + 1][0], b[r + 1][1]));
      ans = max(ans, f[l - 1][0] + next[1] + b[r + 1][1]);
      ans = max(ans, max(f[l - 1][0], f[l - 1][1]) + c[1] + b[r + 1][1]);
      prev[0] = next[0];
      prev[1] = next[1];
    }
  }
  cout << ans << '\n';
  return 0;
}

```