#include <bits/stdc++.h>
using namespace std;
long long n, ans, a[2005];
long long dp[2005][2005][5];
int main() {
  cin.sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
    --a[i];
  }
  for (int i = 1; i <= n; ++i) {
    if (a[i] == 0) {
      dp[i][i][0] = dp[i][i][2] = dp[i][i][3] = 1;
    } else {
      dp[i][i][1] = dp[i][i][2] = dp[i][i][3] = 1;
    }
    for (int j = i + 1; j <= n; ++j) {
      for (int k = 0; k <= 3; ++k) {
        dp[i][j][k] = dp[i][j - 1][k];
      }
      if (a[j] == 0) {
        dp[i][j][0] += 1;
        dp[i][j][2] = max(dp[i][j][2], dp[i][j][0]);
        dp[i][j][3] = max(dp[i][j][3] + 1, dp[i][j][1] + 1);
      } else {
        dp[i][j][1] += 1;
        dp[i][j][2] = max(dp[i][j][2] + 1, dp[i][j][0] + 1);
        dp[i][j][3] = max(dp[i][j][3], dp[i][j][1]);
      }
    }
  }
  ans = max(max(dp[1][n][0], dp[1][n][1]), dp[1][n][2]);
  for (int i = 1; i <= n; ++i) {
    for (int j = i + 1; j <= n; ++j) {
      long long L1 = 1;
      long long R1 = i - 1;
      long long L2 = i;
      long long R2 = j;
      long long L3 = j + 1;
      long long R3 = n;
      ans = max(ans, dp[L1][R1][0] + dp[L2][R2][3] + dp[L3][R3][1]);
    }
  }
  cout << ans << '\n';
}

### Prompt

Construct a cpp code solution to the problem outlined:

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 n, ans, a[2005];
long long dp[2005][2005][5];
int main() {
  cin.sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
    --a[i];
  }
  for (int i = 1; i <= n; ++i) {
    if (a[i] == 0) {
      dp[i][i][0] = dp[i][i][2] = dp[i][i][3] = 1;
    } else {
      dp[i][i][1] = dp[i][i][2] = dp[i][i][3] = 1;
    }
    for (int j = i + 1; j <= n; ++j) {
      for (int k = 0; k <= 3; ++k) {
        dp[i][j][k] = dp[i][j - 1][k];
      }
      if (a[j] == 0) {
        dp[i][j][0] += 1;
        dp[i][j][2] = max(dp[i][j][2], dp[i][j][0]);
        dp[i][j][3] = max(dp[i][j][3] + 1, dp[i][j][1] + 1);
      } else {
        dp[i][j][1] += 1;
        dp[i][j][2] = max(dp[i][j][2] + 1, dp[i][j][0] + 1);
        dp[i][j][3] = max(dp[i][j][3], dp[i][j][1]);
      }
    }
  }
  ans = max(max(dp[1][n][0], dp[1][n][1]), dp[1][n][2]);
  for (int i = 1; i <= n; ++i) {
    for (int j = i + 1; j <= n; ++j) {
      long long L1 = 1;
      long long R1 = i - 1;
      long long L2 = i;
      long long R2 = j;
      long long L3 = j + 1;
      long long R3 = n;
      ans = max(ans, dp[L1][R1][0] + dp[L2][R2][3] + dp[L3][R3][1]);
    }
  }
  cout << ans << '\n';
}

```