#include <bits/stdc++.h>
using namespace std;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  vector<int> a(n);
  vector<vector<int>> pref(2, vector<int>(n, 0));
  for (int &e : a) {
    cin >> e;
    e--;
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < 2; j++) {
      pref[j][i] = (a[i] == j) + (i == 0 ? 0 : pref[j][i - 1]);
    }
  }
  auto calc = [&](int id, int l, int r) {
    return pref[id][r] - (l == 0 ? 0 : pref[id][l - 1]);
  };
  vector<vector<int>> mx(n, vector<int>(n, 0));
  for (int i = 0; i < n; i++) {
    mx[i][i] = 1;
    for (int j = i - 1; j >= 0; j--) {
      mx[j][i] = max(calc(0, j, i), mx[j + 1][i] + (a[j] == 1));
    }
  }
  int ans = 0;
  for (int l = 0; l < n; l++) {
    for (int r = l; r < n; r++) {
      int tmpAns = mx[l][r];
      if (l > 0) tmpAns += calc(0, 0, l - 1);
      if (r != n - 1) tmpAns += calc(1, r + 1, n - 1);
      ans = max(ans, tmpAns);
    }
  }
  cout << ans << '\n';
  return 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::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  vector<int> a(n);
  vector<vector<int>> pref(2, vector<int>(n, 0));
  for (int &e : a) {
    cin >> e;
    e--;
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < 2; j++) {
      pref[j][i] = (a[i] == j) + (i == 0 ? 0 : pref[j][i - 1]);
    }
  }
  auto calc = [&](int id, int l, int r) {
    return pref[id][r] - (l == 0 ? 0 : pref[id][l - 1]);
  };
  vector<vector<int>> mx(n, vector<int>(n, 0));
  for (int i = 0; i < n; i++) {
    mx[i][i] = 1;
    for (int j = i - 1; j >= 0; j--) {
      mx[j][i] = max(calc(0, j, i), mx[j + 1][i] + (a[j] == 1));
    }
  }
  int ans = 0;
  for (int l = 0; l < n; l++) {
    for (int r = l; r < n; r++) {
      int tmpAns = mx[l][r];
      if (l > 0) tmpAns += calc(0, 0, l - 1);
      if (r != n - 1) tmpAns += calc(1, r + 1, n - 1);
      ans = max(ans, tmpAns);
    }
  }
  cout << ans << '\n';
  return 0;
}

```