#include <bits/stdc++.h>
using namespace std;
const int N = 2e3 + 50;
int a[N];
int t1[N];
int t2[N];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    t1[i] = t1[i - 1] + (a[i] == 1);
  }
  t2[n + 1] = 0;
  for (int i = n; i >= 1; i--) {
    t2[i] = t2[i + 1] + (a[i] == 2);
  }
  int ans = 0;
  for (int l = 1; l <= n; l++) {
    int x = t1[l - 1], xx;
    int y = 0, z = 0;
    for (int r = l; r <= n; r++) {
      xx = t2[r + 1];
      if (a[r] == 1)
        y++;
      else {
        z++;
        y = max(z, y);
      }
      ans = max(ans, x + xx + y);
    }
  }
  printf("%d\n", ans);
  return 0;
}

### Prompt

Your task is to create 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 N = 2e3 + 50;
int a[N];
int t1[N];
int t2[N];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    t1[i] = t1[i - 1] + (a[i] == 1);
  }
  t2[n + 1] = 0;
  for (int i = n; i >= 1; i--) {
    t2[i] = t2[i + 1] + (a[i] == 2);
  }
  int ans = 0;
  for (int l = 1; l <= n; l++) {
    int x = t1[l - 1], xx;
    int y = 0, z = 0;
    for (int r = l; r <= n; r++) {
      xx = t2[r + 1];
      if (a[r] == 1)
        y++;
      else {
        z++;
        y = max(z, y);
      }
      ans = max(ans, x + xx + y);
    }
  }
  printf("%d\n", ans);
  return 0;
}

```