#include <bits/stdc++.h>
using namespace std;
const double PI = acos(-1.0);
const double EPS = 1e-9;
const int N = (1 << 15);
const long long OO = 2ll * 1e18;
int a[2005], b[2005], c[2005];
int main() {
  int n, ans = 0;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    b[i] = b[i - 1] + (a[i] == 1 ? 1 : 0);
  }
  for (int i = n; i >= 1; i--) c[i] = c[i + 1] + (a[i] == 2 ? 1 : 0);
  if (n <= 3 || b[n] == 0 || c[1] == 0) {
    cout << n;
    return 0;
  }
  for (int i = 1; i <= n; i++) {
    int cn = 0, mx = 0, mxi = i + 1, mn = 0, mni = i - 1;
    for (int j = i + 1; j <= n; j++) {
      if (a[j] == 1) {
        if (mx < ++cn) {
          mx = cn;
          mxi = j;
        }
      } else
        cn--;
    }
    cn = 0;
    for (int j = i - 1; j >= 1; j--) {
      if (a[j] == 2) {
        if (mn < ++cn) {
          mn = cn;
          mni = j;
        }
      } else
        cn--;
    }
    ans = max(ans, b[mni] + c[mxi] + b[mxi] - b[i - 1] + c[mni] - c[i + 1]);
  }
  cout << ans;
  return 0;
}

### Prompt

In CPP, your task is to solve 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 double PI = acos(-1.0);
const double EPS = 1e-9;
const int N = (1 << 15);
const long long OO = 2ll * 1e18;
int a[2005], b[2005], c[2005];
int main() {
  int n, ans = 0;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    b[i] = b[i - 1] + (a[i] == 1 ? 1 : 0);
  }
  for (int i = n; i >= 1; i--) c[i] = c[i + 1] + (a[i] == 2 ? 1 : 0);
  if (n <= 3 || b[n] == 0 || c[1] == 0) {
    cout << n;
    return 0;
  }
  for (int i = 1; i <= n; i++) {
    int cn = 0, mx = 0, mxi = i + 1, mn = 0, mni = i - 1;
    for (int j = i + 1; j <= n; j++) {
      if (a[j] == 1) {
        if (mx < ++cn) {
          mx = cn;
          mxi = j;
        }
      } else
        cn--;
    }
    cn = 0;
    for (int j = i - 1; j >= 1; j--) {
      if (a[j] == 2) {
        if (mn < ++cn) {
          mn = cn;
          mni = j;
        }
      } else
        cn--;
    }
    ans = max(ans, b[mni] + c[mxi] + b[mxi] - b[i - 1] + c[mni] - c[i + 1]);
  }
  cout << ans;
  return 0;
}

```