#include <bits/stdc++.h>
using namespace std;
long long n, i, a[2020], d1[2020][2020], d2[2020][2020], j, hs;
long long dI(long long aa, long long bb) {
  if (aa > bb) return 0;
  if (aa == bb) return 1;
  if (d1[aa][bb] == -1) {
    if (a[aa] == 1)
      d1[aa][bb] = 1 + dI(aa + 1, bb);
    else if (a[bb] == 2)
      d1[aa][bb] = 1 + dI(aa, bb - 1);
    else
      d1[aa][bb] = max(dI(aa + 1, bb), dI(aa, bb - 1));
  }
  return d1[aa][bb];
}
long long dD(long long aa, long long bb) {
  if (aa > bb) return 0;
  if (aa == bb) return 1;
  if (d2[aa][bb] == -1) {
    if (a[aa] == 2)
      d2[aa][bb] = 1 + dD(aa + 1, bb);
    else if (a[bb] == 1)
      d2[aa][bb] = 1 + dD(aa, bb - 1);
    else
      d2[aa][bb] = max(dD(aa + 1, bb), dD(aa, bb - 1));
  }
  return d2[aa][bb];
}
int main() {
  cin >> n;
  for (i = 1; i <= n; i++) cin >> a[i];
  memset(d1, -1, sizeof(d1));
  memset(d2, -1, sizeof(d2));
  for (i = 1; i <= n; i++) {
    for (j = i; j <= n; j++) {
      hs = max(hs, dI(1, i - 1) + dD(i, j) + dI(j + 1, n));
    }
  }
  cout << hs << "\n";
}

### 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;
long long n, i, a[2020], d1[2020][2020], d2[2020][2020], j, hs;
long long dI(long long aa, long long bb) {
  if (aa > bb) return 0;
  if (aa == bb) return 1;
  if (d1[aa][bb] == -1) {
    if (a[aa] == 1)
      d1[aa][bb] = 1 + dI(aa + 1, bb);
    else if (a[bb] == 2)
      d1[aa][bb] = 1 + dI(aa, bb - 1);
    else
      d1[aa][bb] = max(dI(aa + 1, bb), dI(aa, bb - 1));
  }
  return d1[aa][bb];
}
long long dD(long long aa, long long bb) {
  if (aa > bb) return 0;
  if (aa == bb) return 1;
  if (d2[aa][bb] == -1) {
    if (a[aa] == 2)
      d2[aa][bb] = 1 + dD(aa + 1, bb);
    else if (a[bb] == 1)
      d2[aa][bb] = 1 + dD(aa, bb - 1);
    else
      d2[aa][bb] = max(dD(aa + 1, bb), dD(aa, bb - 1));
  }
  return d2[aa][bb];
}
int main() {
  cin >> n;
  for (i = 1; i <= n; i++) cin >> a[i];
  memset(d1, -1, sizeof(d1));
  memset(d2, -1, sizeof(d2));
  for (i = 1; i <= n; i++) {
    for (j = i; j <= n; j++) {
      hs = max(hs, dI(1, i - 1) + dD(i, j) + dI(j + 1, n));
    }
  }
  cout << hs << "\n";
}

```