#include <bits/stdc++.h>
using namespace std;
int s1[2001];
int s2[2001];
int s21[2000][2000];
int n;
int a[2000];
void Calc21() {
  for (int i = 0; i < n; i++) {
    s21[i][i] = 1;
    int t2 = 0;
    int t1 = 0;
    for (int j = i; j < n; j++) {
      if (a[j] == 2)
        ++t2;
      else
        t1 = max(t2 + 1, t1 + 1);
      s21[i][j] = max(t2, t1);
    }
  }
}
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < n; i++) {
    int q = i;
    while (q >= 0) s1[i + 1] += a[q--] == 1;
  }
  for (int i = 0; i < n; i++) {
    int q = i;
    while (q < n) s2[i] += a[q++] == 2;
  }
  Calc21();
  int best = s21[0][n - 1];
  for (int i = 0; i <= n; i++)
    for (int j = i + 1; j <= n; j++) {
      int cur = s1[i] + s21[i][j - 1];
      cur += s2[j];
      best = max(best, cur);
    }
  cout << best;
  return 0;
}

### 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 s1[2001];
int s2[2001];
int s21[2000][2000];
int n;
int a[2000];
void Calc21() {
  for (int i = 0; i < n; i++) {
    s21[i][i] = 1;
    int t2 = 0;
    int t1 = 0;
    for (int j = i; j < n; j++) {
      if (a[j] == 2)
        ++t2;
      else
        t1 = max(t2 + 1, t1 + 1);
      s21[i][j] = max(t2, t1);
    }
  }
}
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < n; i++) {
    int q = i;
    while (q >= 0) s1[i + 1] += a[q--] == 1;
  }
  for (int i = 0; i < n; i++) {
    int q = i;
    while (q < n) s2[i] += a[q++] == 2;
  }
  Calc21();
  int best = s21[0][n - 1];
  for (int i = 0; i <= n; i++)
    for (int j = i + 1; j <= n; j++) {
      int cur = s1[i] + s21[i][j - 1];
      cur += s2[j];
      best = max(best, cur);
    }
  cout << best;
  return 0;
}

```