#include <bits/stdc++.h>
using namespace ::std;
long long int p[3000];
long long int dp[5][3000];
long long int maxi(long long int a, long long int b) { return a <= b ? b : a; }
int main() {
  long long int i, j, k, n;
  cin >> n;
  long long int ans = 0;
  for (i = 1; i <= n; i++) {
    cin >> p[i];
    if (p[i] == 2) ans++;
  }
  long long int val1 = 0, val2 = 0, val3 = 0, val4 = 0;
  for (i = 1; i <= n; i++) {
    if (p[i] == 1)
      dp[1][i] = dp[1][i - 1] + 1;
    else
      dp[1][i] = dp[1][i - 1];
  }
  for (i = 1; i <= n; i++) {
    if (p[i] == 2) {
      long long int val = maxi(dp[2][i - 1], dp[1][i - 1]);
      dp[2][i] = maxi(dp[2][i - 1], dp[1][i - 1]) + 1;
    } else
      dp[2][i] = dp[2][i - 1];
  }
  for (i = 1; i <= n; i++) {
    if (p[i] == 1) {
      long long int val = maxi(dp[2][i - 1], dp[3][i - 1]);
      dp[3][i] = val + 1;
    } else
      dp[3][i] = dp[3][i - 1];
  }
  for (i = 1; i <= n; i++) {
    if (p[i] == 2) {
      long long int val = maxi(dp[3][i - 1], dp[4][i - 1]);
      dp[4][i] = val + 1;
    } else
      dp[4][i] = dp[4][i - 1];
  }
  ans = maxi(dp[4][n], dp[3][n]);
  ans = maxi(ans, maxi(dp[1][n], dp[2][n]));
  cout << ans << endl;
}

### Prompt

Please formulate 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 int p[3000];
long long int dp[5][3000];
long long int maxi(long long int a, long long int b) { return a <= b ? b : a; }
int main() {
  long long int i, j, k, n;
  cin >> n;
  long long int ans = 0;
  for (i = 1; i <= n; i++) {
    cin >> p[i];
    if (p[i] == 2) ans++;
  }
  long long int val1 = 0, val2 = 0, val3 = 0, val4 = 0;
  for (i = 1; i <= n; i++) {
    if (p[i] == 1)
      dp[1][i] = dp[1][i - 1] + 1;
    else
      dp[1][i] = dp[1][i - 1];
  }
  for (i = 1; i <= n; i++) {
    if (p[i] == 2) {
      long long int val = maxi(dp[2][i - 1], dp[1][i - 1]);
      dp[2][i] = maxi(dp[2][i - 1], dp[1][i - 1]) + 1;
    } else
      dp[2][i] = dp[2][i - 1];
  }
  for (i = 1; i <= n; i++) {
    if (p[i] == 1) {
      long long int val = maxi(dp[2][i - 1], dp[3][i - 1]);
      dp[3][i] = val + 1;
    } else
      dp[3][i] = dp[3][i - 1];
  }
  for (i = 1; i <= n; i++) {
    if (p[i] == 2) {
      long long int val = maxi(dp[3][i - 1], dp[4][i - 1]);
      dp[4][i] = val + 1;
    } else
      dp[4][i] = dp[4][i - 1];
  }
  ans = maxi(dp[4][n], dp[3][n]);
  ans = maxi(ans, maxi(dp[1][n], dp[2][n]));
  cout << ans << endl;
}

```