#include <bits/stdc++.h>
using namespace std;
const int MAX = 2e3 + 1;
int dp[4], in[MAX], mx[4];
int main() {
  ios::sync_with_stdio(0);
  int n, ans = 1;
  cin >> n;
  for (int i = 0; i < n; ++i) {
    cin >> in[i];
    if (in[i] == 1) {
      dp[0] = 1;
      dp[0] = max(dp[0], mx[0] + 1);
      dp[2] = max(dp[2], mx[2] + 1);
      dp[2] = max(dp[2], mx[1] + 1);
      mx[0] = max(mx[0], dp[0]);
      mx[2] = max(mx[2], dp[2]);
    }
    if (in[i] == 2) {
      dp[1] = 1;
      dp[1] = max(dp[1], mx[1] + 1);
      dp[3] = max(dp[3], mx[3] + 1);
      dp[1] = max(dp[1], mx[0] + 1);
      dp[3] = max(dp[3], mx[2] + 1);
      mx[1] = max(mx[1], dp[1]);
      mx[3] = max(mx[3], dp[3]);
    }
    for (int j = 0; j < 4; ++j) {
      ans = max(ans, dp[j]);
      dp[j] = 0;
    }
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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 MAX = 2e3 + 1;
int dp[4], in[MAX], mx[4];
int main() {
  ios::sync_with_stdio(0);
  int n, ans = 1;
  cin >> n;
  for (int i = 0; i < n; ++i) {
    cin >> in[i];
    if (in[i] == 1) {
      dp[0] = 1;
      dp[0] = max(dp[0], mx[0] + 1);
      dp[2] = max(dp[2], mx[2] + 1);
      dp[2] = max(dp[2], mx[1] + 1);
      mx[0] = max(mx[0], dp[0]);
      mx[2] = max(mx[2], dp[2]);
    }
    if (in[i] == 2) {
      dp[1] = 1;
      dp[1] = max(dp[1], mx[1] + 1);
      dp[3] = max(dp[3], mx[3] + 1);
      dp[1] = max(dp[1], mx[0] + 1);
      dp[3] = max(dp[3], mx[2] + 1);
      mx[1] = max(mx[1], dp[1]);
      mx[3] = max(mx[3], dp[3]);
    }
    for (int j = 0; j < 4; ++j) {
      ans = max(ans, dp[j]);
      dp[j] = 0;
    }
  }
  cout << ans << endl;
  return 0;
}

```