#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  cin >> n;
  int a[n];
  for (long long i = 0; i < n; i++) cin >> a[i];
  vector<vector<int> > lds(n, vector<int>(n, 0)), twos(n, vector<int>(n, 0));
  for (long long i = 0; i < n; i++) {
    twos[i][i] = (a[i] == 2);
    lds[i][i] = 1;
    for (long long j = i + 1; j < n; j++) {
      twos[i][j] = twos[i][j - 1] + (a[j] == 2);
      lds[i][j] = max(lds[i][j - 1], twos[i][j]);
      if (a[j] == 1) lds[i][j] = max(lds[i][j], lds[i][j - 1] + 1);
    }
  }
  int lOnes[n], rTwos[n];
  int x = 0;
  for (long long i = 0; i < n; i++) {
    if (a[i] == 1) ++x;
    lOnes[i] = x;
  }
  x = 0;
  for (long long i = n - 1; i >= 0; i--) {
    if (a[i] == 2) ++x;
    rTwos[i] = x;
  }
  int ans = 0;
  for (long long i = 0; i < n; i++) {
    for (long long j = i; j < n; j++) {
      x = 0;
      if (i > 0) x += lOnes[i - 1];
      if (j < n - 1) x += rTwos[j + 1];
      x += lds[i][j];
      ans = max(ans, x);
    }
  }
  cout << ans << endl;
  exit(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;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  cin >> n;
  int a[n];
  for (long long i = 0; i < n; i++) cin >> a[i];
  vector<vector<int> > lds(n, vector<int>(n, 0)), twos(n, vector<int>(n, 0));
  for (long long i = 0; i < n; i++) {
    twos[i][i] = (a[i] == 2);
    lds[i][i] = 1;
    for (long long j = i + 1; j < n; j++) {
      twos[i][j] = twos[i][j - 1] + (a[j] == 2);
      lds[i][j] = max(lds[i][j - 1], twos[i][j]);
      if (a[j] == 1) lds[i][j] = max(lds[i][j], lds[i][j - 1] + 1);
    }
  }
  int lOnes[n], rTwos[n];
  int x = 0;
  for (long long i = 0; i < n; i++) {
    if (a[i] == 1) ++x;
    lOnes[i] = x;
  }
  x = 0;
  for (long long i = n - 1; i >= 0; i--) {
    if (a[i] == 2) ++x;
    rTwos[i] = x;
  }
  int ans = 0;
  for (long long i = 0; i < n; i++) {
    for (long long j = i; j < n; j++) {
      x = 0;
      if (i > 0) x += lOnes[i - 1];
      if (j < n - 1) x += rTwos[j + 1];
      x += lds[i][j];
      ans = max(ans, x);
    }
  }
  cout << ans << endl;
  exit(0);
}

```