#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 7;
long long n, a[N], p[N][2], dp[2];
long long get(int l, int r, int x) { return p[r][x] - p[l - 1][x]; }
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    a[i] -= 1;
    p[i][a[i]] = 1;
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < 2; j++) {
      p[i][j] += p[i - 1][j];
    }
  }
  dp[0] = dp[1] = 0;
  for (int i = 1; i <= n; i++) {
    if (a[i] == 0)
      dp[0]++;
    else {
      dp[1]++;
      dp[1] = max(dp[1], dp[0] + 1);
    }
  }
  long long ans = max(dp[0], dp[1]);
  for (int i = 1; i <= n; i++) {
    memset(dp, 0, sizeof(dp));
    for (int j = i; j <= n; j++) {
      if (a[j] == 1)
        dp[1]++;
      else {
        dp[0]++;
        dp[0] = max(dp[0], dp[1] + 1);
      }
      ans = max(ans, get(1, i - 1, 0) + get(j + 1, n, 1) + max(dp[0], dp[1]));
    }
  }
  cout << ans << '\n';
  return 0;
}

### Prompt

Please create a solution in Cpp 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 = 1e5 + 7;
long long n, a[N], p[N][2], dp[2];
long long get(int l, int r, int x) { return p[r][x] - p[l - 1][x]; }
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    a[i] -= 1;
    p[i][a[i]] = 1;
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < 2; j++) {
      p[i][j] += p[i - 1][j];
    }
  }
  dp[0] = dp[1] = 0;
  for (int i = 1; i <= n; i++) {
    if (a[i] == 0)
      dp[0]++;
    else {
      dp[1]++;
      dp[1] = max(dp[1], dp[0] + 1);
    }
  }
  long long ans = max(dp[0], dp[1]);
  for (int i = 1; i <= n; i++) {
    memset(dp, 0, sizeof(dp));
    for (int j = i; j <= n; j++) {
      if (a[j] == 1)
        dp[1]++;
      else {
        dp[0]++;
        dp[0] = max(dp[0], dp[1] + 1);
      }
      ans = max(ans, get(1, i - 1, 0) + get(j + 1, n, 1) + max(dp[0], dp[1]));
    }
  }
  cout << ans << '\n';
  return 0;
}

```