#include <bits/stdc++.h>
using namespace std;
const double PI = acos(-1);
const long long N = 2e3 + 5;
const long long mod = 1e9 + 7;
const long long INF = 1e18;
long long a[N], endingAt2[N], startingAt1[N];
long long dp[N][N][3];
long long pre1[N], pre2[N];
void solve() {
  long long n;
  cin >> n;
  pre1[0] = pre2[2] = 0;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    if (a[i] == 1)
      pre1[i] = pre1[i - 1] + 1, pre2[i] = pre2[i - 1];
    else
      pre2[i] = pre2[i - 1] + 1, pre1[i] = pre1[i - 1];
  }
  long long ans = max(pre1[n], pre2[n]);
  endingAt2[0] = 0;
  for (int i = 1; i <= n; i++) {
    if (a[i] == 1)
      endingAt2[i] = endingAt2[i - 1];
    else
      endingAt2[i] = max(pre1[i - 1], endingAt2[i - 1]) + 1;
  }
  ans = max(ans, endingAt2[n]);
  startingAt1[n + 1] = 0;
  for (int i = n; i > 0; i--) {
    if (a[i] == 1)
      startingAt1[i] = max(startingAt1[i + 1], (pre2[n] - pre2[i])) + 1;
    else
      startingAt1[i] = startingAt1[i + 1];
  }
  ans = max(ans, startingAt1[1]);
  for (int i = 1; i <= n; i++) {
    dp[i][i - 1][1] = dp[i][i - 1][2] = 0;
    for (int j = i; j <= n; j++) {
      if (a[j] == 1) {
        dp[i][j][1] = max(dp[i][j - 1][1], dp[i][j - 1][2]) + 1;
        dp[i][j][2] = dp[i][j - 1][2];
      } else {
        dp[i][j][1] = dp[i][j - 1][1];
        dp[i][j][2] = dp[i][j - 1][2] + 1;
      }
    }
  }
  for (int i = 1; i <= n; i++) {
    for (int j = i + 1; j <= n; j++) {
      ans = max(ans, pre1[j] + max(startingAt1[j + 1], (pre2[n] - pre2[j])));
      ans = max(ans, pre1[i - 1] + max(dp[i][j][1], dp[i][j][2]) +
                         (pre2[n] - pre2[j]));
      ans = max(ans, pre1[i - 1] + pre2[n] - pre2[i - 1]);
      ans = max(ans, endingAt2[i - 1] + pre2[n] - pre2[i - 1]);
    }
  }
  cout << ans << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  solve();
}

### 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;
const double PI = acos(-1);
const long long N = 2e3 + 5;
const long long mod = 1e9 + 7;
const long long INF = 1e18;
long long a[N], endingAt2[N], startingAt1[N];
long long dp[N][N][3];
long long pre1[N], pre2[N];
void solve() {
  long long n;
  cin >> n;
  pre1[0] = pre2[2] = 0;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    if (a[i] == 1)
      pre1[i] = pre1[i - 1] + 1, pre2[i] = pre2[i - 1];
    else
      pre2[i] = pre2[i - 1] + 1, pre1[i] = pre1[i - 1];
  }
  long long ans = max(pre1[n], pre2[n]);
  endingAt2[0] = 0;
  for (int i = 1; i <= n; i++) {
    if (a[i] == 1)
      endingAt2[i] = endingAt2[i - 1];
    else
      endingAt2[i] = max(pre1[i - 1], endingAt2[i - 1]) + 1;
  }
  ans = max(ans, endingAt2[n]);
  startingAt1[n + 1] = 0;
  for (int i = n; i > 0; i--) {
    if (a[i] == 1)
      startingAt1[i] = max(startingAt1[i + 1], (pre2[n] - pre2[i])) + 1;
    else
      startingAt1[i] = startingAt1[i + 1];
  }
  ans = max(ans, startingAt1[1]);
  for (int i = 1; i <= n; i++) {
    dp[i][i - 1][1] = dp[i][i - 1][2] = 0;
    for (int j = i; j <= n; j++) {
      if (a[j] == 1) {
        dp[i][j][1] = max(dp[i][j - 1][1], dp[i][j - 1][2]) + 1;
        dp[i][j][2] = dp[i][j - 1][2];
      } else {
        dp[i][j][1] = dp[i][j - 1][1];
        dp[i][j][2] = dp[i][j - 1][2] + 1;
      }
    }
  }
  for (int i = 1; i <= n; i++) {
    for (int j = i + 1; j <= n; j++) {
      ans = max(ans, pre1[j] + max(startingAt1[j + 1], (pre2[n] - pre2[j])));
      ans = max(ans, pre1[i - 1] + max(dp[i][j][1], dp[i][j][2]) +
                         (pre2[n] - pre2[j]));
      ans = max(ans, pre1[i - 1] + pre2[n] - pre2[i - 1]);
      ans = max(ans, endingAt2[i - 1] + pre2[n] - pre2[i - 1]);
    }
  }
  cout << ans << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  solve();
}

```