#include <bits/stdc++.h>
using namespace std;
long long INF = 1e18 + 10;
long double eps = 1e-9;
int main() {
  cin.tie(0);
  cin.sync_with_stdio(0);
  cout.tie(0);
  int n;
  cin >> n;
  vector<int> v(n);
  vector<int> pref1(n, 0), last2(n, 0);
  for (int i = 0; i < n; i++) {
    cin >> v[i];
  }
  for (int i = 0; i < n; i++) {
    if (i != 0) pref1[i] = pref1[i - 1];
    if (v[i] == 1) {
      pref1[i]++;
    }
  }
  for (int i = n - 1; i >= 0; i--) {
    if (i != n - 1) last2[i] = last2[i + 1];
    if (v[i] == 2) {
      last2[i]++;
    }
  }
  long long ans = 0;
  ans = max(last2[0], pref1[n - 1]);
  for (int i = 0; i < n - 1; i++) {
    ans = max(pref1[i], last2[i + 1]);
  }
  for (int i = 0; i < n; i++) {
    int d = 0, cnt2 = 0, cnt1 = 0;
    for (int j = i; j < n; j++) {
      cnt1 += (v[j] == 1);
      cnt2 += (v[j] == 2);
      if (cnt2 >= cnt1) {
        d += cnt2;
        cnt1 = 0;
        cnt2 = 0;
      }
      long long sum = d + cnt1;
      if (i != 0) {
        sum += pref1[i - 1];
      }
      if (j != n - 1) {
        sum += last2[j + 1];
      }
      ans = max(ans, sum);
    }
  }
  cout << ans;
  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;
long long INF = 1e18 + 10;
long double eps = 1e-9;
int main() {
  cin.tie(0);
  cin.sync_with_stdio(0);
  cout.tie(0);
  int n;
  cin >> n;
  vector<int> v(n);
  vector<int> pref1(n, 0), last2(n, 0);
  for (int i = 0; i < n; i++) {
    cin >> v[i];
  }
  for (int i = 0; i < n; i++) {
    if (i != 0) pref1[i] = pref1[i - 1];
    if (v[i] == 1) {
      pref1[i]++;
    }
  }
  for (int i = n - 1; i >= 0; i--) {
    if (i != n - 1) last2[i] = last2[i + 1];
    if (v[i] == 2) {
      last2[i]++;
    }
  }
  long long ans = 0;
  ans = max(last2[0], pref1[n - 1]);
  for (int i = 0; i < n - 1; i++) {
    ans = max(pref1[i], last2[i + 1]);
  }
  for (int i = 0; i < n; i++) {
    int d = 0, cnt2 = 0, cnt1 = 0;
    for (int j = i; j < n; j++) {
      cnt1 += (v[j] == 1);
      cnt2 += (v[j] == 2);
      if (cnt2 >= cnt1) {
        d += cnt2;
        cnt1 = 0;
        cnt2 = 0;
      }
      long long sum = d + cnt1;
      if (i != 0) {
        sum += pref1[i - 1];
      }
      if (j != n - 1) {
        sum += last2[j + 1];
      }
      ans = max(ans, sum);
    }
  }
  cout << ans;
  return 0;
}

```