#include <bits/stdc++.h>
using namespace std;
int i, j, k, n, t, f[4], ans, a[2048], l1[2048], r2[2048];
int main() {
  scanf("%d", &n);
  for (i = 1; i <= n; ++i) {
    scanf("%d", a + i);
    l1[i] = l1[i - 1] + (a[i] == 1);
    for (j = a[i]; ~j; --j)
      if (f[j] + 1 > f[a[i]]) f[a[i]] = f[j] + 1;
  }
  ans = f[1] > f[2] ? f[1] : f[2];
  for (i = n; i; --i) r2[i] = r2[i + 1] + (a[i] == 2);
  for (i = 1; i <= n; ++i) {
    f[1] = f[2] = 0;
    for (k = i; k <= n; ++k) {
      for (j = a[k]; j < 4; ++j)
        if (f[j] + 1 > f[a[k]]) f[a[k]] = f[j] + 1;
      t = l1[i - 1] + r2[k + 1] + (f[1] > f[2] ? f[1] : f[2]);
      ans = ans < t ? t : ans;
    }
  }
  printf("%d", ans);
}

### Prompt

Please create a solution in CPP 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;
int i, j, k, n, t, f[4], ans, a[2048], l1[2048], r2[2048];
int main() {
  scanf("%d", &n);
  for (i = 1; i <= n; ++i) {
    scanf("%d", a + i);
    l1[i] = l1[i - 1] + (a[i] == 1);
    for (j = a[i]; ~j; --j)
      if (f[j] + 1 > f[a[i]]) f[a[i]] = f[j] + 1;
  }
  ans = f[1] > f[2] ? f[1] : f[2];
  for (i = n; i; --i) r2[i] = r2[i + 1] + (a[i] == 2);
  for (i = 1; i <= n; ++i) {
    f[1] = f[2] = 0;
    for (k = i; k <= n; ++k) {
      for (j = a[k]; j < 4; ++j)
        if (f[j] + 1 > f[a[k]]) f[a[k]] = f[j] + 1;
      t = l1[i - 1] + r2[k + 1] + (f[1] > f[2] ? f[1] : f[2]);
      ans = ans < t ? t : ans;
    }
  }
  printf("%d", ans);
}

```