#include <bits/stdc++.h>
using namespace std;
const int N = 2005;
int n, a[N], dp[N], dp2[N], c1[N];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < n; i++) {
    c1[i] = a[i] == 1;
    if (i) c1[i] += c1[i - 1];
  }
  for (int i = 0; i < n; i++) {
    dp[i] = 1;
    for (int j = 0; j < i; j++) {
      if (a[i] >= a[j]) dp[i] = max(dp[i], dp[j] + 1);
    }
  }
  for (int i = n - 1; i >= 0; i--) {
    dp2[i] = 1;
    for (int j = i + 1; j < n; j++) {
      if (a[j] >= a[i]) dp2[i] = max(dp2[i], dp2[j] + 1);
    }
  }
  int ans = 0;
  for (int i = 0; i < n; i++) {
    vector<int> v;
    for (int j = i; j < n; j++) {
      if (v.empty())
        v.push_back(-a[j]);
      else if (v.back() <= -a[j]) {
        v.push_back(-a[j]);
      } else {
        auto it = upper_bound(v.begin(), v.end(), -a[j]) - v.begin();
        v[it] = -a[j];
      }
      int c = c1[j];
      if (i) c -= c1[i - 1];
      int c2 = j - i + 1 - c;
      int best = v.size();
      int ans1 = 0;
      if (i) ans1 += dp[i - 1];
      ans1 += c2;
      ans1 += (n - j - 1) - (c1[n - 1] - c1[j]);
      ans = max(ans, ans1);
      int ans2 = 0;
      if (i) ans2 += c1[i - 1];
      ans2 += best;
      ans2 += (n - j - 1) - (c1[n - 1] - c1[j]);
      ans = max(ans, ans2);
      int ans3 = 0;
      ans3 += c1[j];
      if (j != n - 1) ans3 += dp2[j + 1];
      ans = max(ans, ans3);
    }
  }
  assert(ans >= dp[n - 1]);
  assert(ans >= dp2[0]);
  cout << ans;
  return 0;
}

### Prompt

Generate 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;
const int N = 2005;
int n, a[N], dp[N], dp2[N], c1[N];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < n; i++) {
    c1[i] = a[i] == 1;
    if (i) c1[i] += c1[i - 1];
  }
  for (int i = 0; i < n; i++) {
    dp[i] = 1;
    for (int j = 0; j < i; j++) {
      if (a[i] >= a[j]) dp[i] = max(dp[i], dp[j] + 1);
    }
  }
  for (int i = n - 1; i >= 0; i--) {
    dp2[i] = 1;
    for (int j = i + 1; j < n; j++) {
      if (a[j] >= a[i]) dp2[i] = max(dp2[i], dp2[j] + 1);
    }
  }
  int ans = 0;
  for (int i = 0; i < n; i++) {
    vector<int> v;
    for (int j = i; j < n; j++) {
      if (v.empty())
        v.push_back(-a[j]);
      else if (v.back() <= -a[j]) {
        v.push_back(-a[j]);
      } else {
        auto it = upper_bound(v.begin(), v.end(), -a[j]) - v.begin();
        v[it] = -a[j];
      }
      int c = c1[j];
      if (i) c -= c1[i - 1];
      int c2 = j - i + 1 - c;
      int best = v.size();
      int ans1 = 0;
      if (i) ans1 += dp[i - 1];
      ans1 += c2;
      ans1 += (n - j - 1) - (c1[n - 1] - c1[j]);
      ans = max(ans, ans1);
      int ans2 = 0;
      if (i) ans2 += c1[i - 1];
      ans2 += best;
      ans2 += (n - j - 1) - (c1[n - 1] - c1[j]);
      ans = max(ans, ans2);
      int ans3 = 0;
      ans3 += c1[j];
      if (j != n - 1) ans3 += dp2[j + 1];
      ans = max(ans, ans3);
    }
  }
  assert(ans >= dp[n - 1]);
  assert(ans >= dp2[0]);
  cout << ans;
  return 0;
}

```