#include <bits/stdc++.h>
using namespace std;
int main() {
  int n, i, j, ans, a1, a2, ma2, mb1, b1, b2, A[1009];
  scanf("%d", &n);
  for (i = 0; i < n; ++i) scanf("%d", A + i);
  for (i = 0, ans = 0; i < n; ans = max(ans, a1 + b2 + max(ma2, mb1)), ++i) {
    for (j = 0, a1 = 0, a2 = 0, ma2 = 0; j < i; ++j) {
      if (A[j] == 2)
        ma2 = max(ma2, ++a2);
      else
        ++a1, a2 = max(--a2, 0);
    }
    for (mb1 = 0, b1 = 0, b2 = 0; j < n; ++j) {
      if (A[j] == 1)
        mb1 = max(mb1, ++b1);
      else
        ++b2, b1 = max(--b1, 0);
    }
  }
  printf("%d\n", ans), exit(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;
int main() {
  int n, i, j, ans, a1, a2, ma2, mb1, b1, b2, A[1009];
  scanf("%d", &n);
  for (i = 0; i < n; ++i) scanf("%d", A + i);
  for (i = 0, ans = 0; i < n; ans = max(ans, a1 + b2 + max(ma2, mb1)), ++i) {
    for (j = 0, a1 = 0, a2 = 0, ma2 = 0; j < i; ++j) {
      if (A[j] == 2)
        ma2 = max(ma2, ++a2);
      else
        ++a1, a2 = max(--a2, 0);
    }
    for (mb1 = 0, b1 = 0, b2 = 0; j < n; ++j) {
      if (A[j] == 1)
        mb1 = max(mb1, ++b1);
      else
        ++b2, b1 = max(--b1, 0);
    }
  }
  printf("%d\n", ans), exit(0);
}

```