#include <bits/stdc++.h>
using namespace std;
int dp[2005][2];
int rdp[2005][2];
int fdp[2005][2005][2];
int pref[2005][2];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  int n;
  cin >> n;
  int arr[n + 1];
  for (int i = 1; i < n + 1; i++) {
    cin >> arr[i];
    if (arr[i] == 1) {
      dp[i][0] = dp[i - 1][0] + 1;
      dp[i][1] = dp[i - 1][1];
      pref[i][1] = pref[i - 1][1];
      pref[i][0] = pref[i - 1][0] + 1;
    } else {
      dp[i][1] = max(dp[i - 1][0], dp[i - 1][1]) + 1;
      dp[i][0] = dp[i - 1][0];
      pref[i][1] = pref[i - 1][1] + 1;
      pref[i][0] = pref[i - 1][0];
    }
  }
  for (int i = n + 1 - 1; i >= 1; i--) {
    if (arr[i] == 2) {
      rdp[i][0] = rdp[i + 1][0];
      rdp[i][1] = rdp[i + 1][1] + 1;
    } else {
      rdp[i][0] = max(rdp[i + 1][0], rdp[i + 1][1]) + 1;
      rdp[i][1] = rdp[i + 1][1];
    }
  }
  int ans = max(dp[n][0], dp[n][1]);
  for (int j = 1; j < n + 1; j++) {
    for (int i = j; i < n + 1; i++) {
      if (arr[i] == 2) {
        fdp[j][i][0] = fdp[j][i - 1][0] + 1;
        fdp[j][i][1] = fdp[j][i - 1][1];
      } else {
        fdp[j][i][1] = max(fdp[j][i - 1][0], fdp[j][i - 1][1]) + 1;
        fdp[j][i][0] = fdp[j][i - 1][0];
      }
    }
  }
  for (int i = 1; i < n + 1; i++) {
    for (int j = i + 1; j < n + 1; j++) {
      int t1 = dp[i - 1][0];
      int t2 = dp[i - 1][1];
      int rt1 = rdp[j + 1][0];
      int rt2 = rdp[j + 1][1];
      int p1 = pref[j][0] - pref[i - 1][0];
      int p2 = pref[j][1] - pref[i - 1][1];
      ans = max(ans, t1 + p1 + max(rt1, rt2));
      ans = max(ans, t2 + rt2 + p2);
      ans = max(ans, t1 + rt2 + max(fdp[i][j][0], fdp[i][j][1]));
    }
  }
  cout << ans;
}

### Prompt

Your challenge is to write 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 dp[2005][2];
int rdp[2005][2];
int fdp[2005][2005][2];
int pref[2005][2];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  int n;
  cin >> n;
  int arr[n + 1];
  for (int i = 1; i < n + 1; i++) {
    cin >> arr[i];
    if (arr[i] == 1) {
      dp[i][0] = dp[i - 1][0] + 1;
      dp[i][1] = dp[i - 1][1];
      pref[i][1] = pref[i - 1][1];
      pref[i][0] = pref[i - 1][0] + 1;
    } else {
      dp[i][1] = max(dp[i - 1][0], dp[i - 1][1]) + 1;
      dp[i][0] = dp[i - 1][0];
      pref[i][1] = pref[i - 1][1] + 1;
      pref[i][0] = pref[i - 1][0];
    }
  }
  for (int i = n + 1 - 1; i >= 1; i--) {
    if (arr[i] == 2) {
      rdp[i][0] = rdp[i + 1][0];
      rdp[i][1] = rdp[i + 1][1] + 1;
    } else {
      rdp[i][0] = max(rdp[i + 1][0], rdp[i + 1][1]) + 1;
      rdp[i][1] = rdp[i + 1][1];
    }
  }
  int ans = max(dp[n][0], dp[n][1]);
  for (int j = 1; j < n + 1; j++) {
    for (int i = j; i < n + 1; i++) {
      if (arr[i] == 2) {
        fdp[j][i][0] = fdp[j][i - 1][0] + 1;
        fdp[j][i][1] = fdp[j][i - 1][1];
      } else {
        fdp[j][i][1] = max(fdp[j][i - 1][0], fdp[j][i - 1][1]) + 1;
        fdp[j][i][0] = fdp[j][i - 1][0];
      }
    }
  }
  for (int i = 1; i < n + 1; i++) {
    for (int j = i + 1; j < n + 1; j++) {
      int t1 = dp[i - 1][0];
      int t2 = dp[i - 1][1];
      int rt1 = rdp[j + 1][0];
      int rt2 = rdp[j + 1][1];
      int p1 = pref[j][0] - pref[i - 1][0];
      int p2 = pref[j][1] - pref[i - 1][1];
      ans = max(ans, t1 + p1 + max(rt1, rt2));
      ans = max(ans, t2 + rt2 + p2);
      ans = max(ans, t1 + rt2 + max(fdp[i][j][0], fdp[i][j][1]));
    }
  }
  cout << ans;
}

```