#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 7;
const long long inf = (1ll << 62);
const int MX = 2009;
int n, a[MX], pre[MX][4], suf[MX][4], mx[MX][MX][3][3];
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
  }
  pre[1][a[1]] = 1;
  for (int i = 2; i <= n; i++) {
    if (a[i] == 1) {
      pre[i][2] = pre[i - 1][2];
      pre[i][1] = pre[i - 1][1] + 1;
      continue;
    }
    pre[i][1] = pre[i - 1][1];
    pre[i][2] = max(pre[i - 1][2] + 1, pre[i - 1][1] + 1);
  }
  suf[n][a[n]] = 1;
  for (int i = n - 1; i >= 1; i--) {
    if (a[i] == 2) {
      suf[i][1] = suf[i + 1][1];
      suf[i][2] = suf[i + 1][2] + 1;
      continue;
    }
    suf[i][2] = suf[i + 1][2];
    suf[i][1] = max(suf[i + 1][1] + 1, suf[i + 1][2] + 1);
  }
  for (int i = n; i >= 1; i--) {
    int ones = 0, twos = 0, twone = 0;
    for (int j = i; j >= 1; j--) {
      if (a[j] == 1)
        ones++;
      else {
        twos++;
        if (ones) {
          twone = max(ones + 1, twone + 1);
        }
      }
      mx[j][i][1][1] = ones;
      mx[j][i][2][2] = twos;
      mx[j][i][1][2] = twone;
    }
  }
  int ans = 1;
  for (int i = 1; i <= n; i++) {
    for (int j = i; j <= n; j++) {
      ans = max(ans, pre[i - 1][1] + mx[i][j][1][1] + suf[j + 1][1]);
      ans = max(ans, pre[i - 1][1] + mx[i][j][1][1] + suf[j + 1][2]);
      ans = max(ans, pre[i - 1][1] + mx[i][j][1][2] + suf[j + 1][2]);
      ans = max(ans, pre[i - 1][1] + mx[i][j][2][2] + suf[j + 1][2]);
      ans = max(ans, pre[i - 1][2] + mx[i][j][2][2] + suf[j + 1][2]);
    }
  }
  ans = max(ans, pre[n][1]);
  ans = max(ans, pre[n][2]);
  cout << ans << endl;
}

### Prompt

Develop a solution in cpp to the problem described below:

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 long long mod = 1e9 + 7;
const long long inf = (1ll << 62);
const int MX = 2009;
int n, a[MX], pre[MX][4], suf[MX][4], mx[MX][MX][3][3];
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
  }
  pre[1][a[1]] = 1;
  for (int i = 2; i <= n; i++) {
    if (a[i] == 1) {
      pre[i][2] = pre[i - 1][2];
      pre[i][1] = pre[i - 1][1] + 1;
      continue;
    }
    pre[i][1] = pre[i - 1][1];
    pre[i][2] = max(pre[i - 1][2] + 1, pre[i - 1][1] + 1);
  }
  suf[n][a[n]] = 1;
  for (int i = n - 1; i >= 1; i--) {
    if (a[i] == 2) {
      suf[i][1] = suf[i + 1][1];
      suf[i][2] = suf[i + 1][2] + 1;
      continue;
    }
    suf[i][2] = suf[i + 1][2];
    suf[i][1] = max(suf[i + 1][1] + 1, suf[i + 1][2] + 1);
  }
  for (int i = n; i >= 1; i--) {
    int ones = 0, twos = 0, twone = 0;
    for (int j = i; j >= 1; j--) {
      if (a[j] == 1)
        ones++;
      else {
        twos++;
        if (ones) {
          twone = max(ones + 1, twone + 1);
        }
      }
      mx[j][i][1][1] = ones;
      mx[j][i][2][2] = twos;
      mx[j][i][1][2] = twone;
    }
  }
  int ans = 1;
  for (int i = 1; i <= n; i++) {
    for (int j = i; j <= n; j++) {
      ans = max(ans, pre[i - 1][1] + mx[i][j][1][1] + suf[j + 1][1]);
      ans = max(ans, pre[i - 1][1] + mx[i][j][1][1] + suf[j + 1][2]);
      ans = max(ans, pre[i - 1][1] + mx[i][j][1][2] + suf[j + 1][2]);
      ans = max(ans, pre[i - 1][1] + mx[i][j][2][2] + suf[j + 1][2]);
      ans = max(ans, pre[i - 1][2] + mx[i][j][2][2] + suf[j + 1][2]);
    }
  }
  ans = max(ans, pre[n][1]);
  ans = max(ans, pre[n][2]);
  cout << ans << endl;
}

```