#include <bits/stdc++.h>
using namespace std;
int a1[2005], a2[2005], dp[2005][2005];
int main() {
  int n;
  cin >> n;
  vector<int> a, v;
  int now;
  int t1 = 0, t2 = 0;
  for (int i = 0; i < n; ++i) {
    cin >> now;
    a.push_back(now);
    if (now == 1)
      t1++;
    else
      t2++;
  }
  int t11 = 0, t22 = t2;
  for (int i = 0; i < n; ++i) {
    if (a[i] == 1) t11++;
    a1[i] = t11;
    a2[i] = t22;
    if (a[i] == 2) t22--;
  }
  for (int i = 0; i < a.size(); ++i) dp[i][i] = 1;
  for (int i = a.size() - 1; i >= 0; --i) {
    for (int j = i - 1; j >= 0; --j) {
      if (a[j] == 1) {
        dp[j][i] = max(dp[j + 1][i], a1[i] - a1[j] + 1);
      } else {
        dp[j][i] = dp[j + 1][i] + 1;
      }
    }
  }
  int mx = 0;
  for (int i = 0; i < a.size(); ++i) {
    for (int j = i; j < a.size(); ++j) {
      int now = 0;
      if (i - 1 >= 0) now += a1[i - 1];
      if (j + 1 < a.size()) now += a2[j + 1];
      now += dp[i][j];
      mx = max(mx, now);
    }
  }
  cout << mx << endl;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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;
int a1[2005], a2[2005], dp[2005][2005];
int main() {
  int n;
  cin >> n;
  vector<int> a, v;
  int now;
  int t1 = 0, t2 = 0;
  for (int i = 0; i < n; ++i) {
    cin >> now;
    a.push_back(now);
    if (now == 1)
      t1++;
    else
      t2++;
  }
  int t11 = 0, t22 = t2;
  for (int i = 0; i < n; ++i) {
    if (a[i] == 1) t11++;
    a1[i] = t11;
    a2[i] = t22;
    if (a[i] == 2) t22--;
  }
  for (int i = 0; i < a.size(); ++i) dp[i][i] = 1;
  for (int i = a.size() - 1; i >= 0; --i) {
    for (int j = i - 1; j >= 0; --j) {
      if (a[j] == 1) {
        dp[j][i] = max(dp[j + 1][i], a1[i] - a1[j] + 1);
      } else {
        dp[j][i] = dp[j + 1][i] + 1;
      }
    }
  }
  int mx = 0;
  for (int i = 0; i < a.size(); ++i) {
    for (int j = i; j < a.size(); ++j) {
      int now = 0;
      if (i - 1 >= 0) now += a1[i - 1];
      if (j + 1 < a.size()) now += a2[j + 1];
      now += dp[i][j];
      mx = max(mx, now);
    }
  }
  cout << mx << endl;
}

```