#include <bits/stdc++.h>
using namespace std;
const int maxn = 2005;
const int Mod = 1000000007;
const int INF = 0x3f3f3f3f;
const long long LL_INF = 0x3f3f3f3f3f3f3f3f;
const double PI = acos(-1);
const double ERR = 1e-10;
int n, a[maxn], dp1[maxn], dp2[maxn];
int maxx, l, r;
int main() {
  maxx = 0;
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i <= n; i++) dp1[i] = dp1[i - 1] + (a[i] == 1);
  for (int i = n; i > 0; i--) dp2[i] = dp2[i + 1] + (a[i] == 2);
  for (int i = 1; i <= n; i++) maxx = max(maxx, dp1[i] + dp2[i]);
  for (int i = 1; i <= n; i++) {
    l = 0, r = 0;
    for (int j = i; j <= n; j++) {
      if (a[j] == 1)
        l++;
      else {
        r++;
        l--;
        if (l < 0) l = 0;
      }
      maxx = max(maxx, dp1[i - 1] + l + r + dp2[j + 1]);
    }
  }
  cout << maxx << endl;
  return 0;
}

### Prompt

Generate 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 int maxn = 2005;
const int Mod = 1000000007;
const int INF = 0x3f3f3f3f;
const long long LL_INF = 0x3f3f3f3f3f3f3f3f;
const double PI = acos(-1);
const double ERR = 1e-10;
int n, a[maxn], dp1[maxn], dp2[maxn];
int maxx, l, r;
int main() {
  maxx = 0;
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i <= n; i++) dp1[i] = dp1[i - 1] + (a[i] == 1);
  for (int i = n; i > 0; i--) dp2[i] = dp2[i + 1] + (a[i] == 2);
  for (int i = 1; i <= n; i++) maxx = max(maxx, dp1[i] + dp2[i]);
  for (int i = 1; i <= n; i++) {
    l = 0, r = 0;
    for (int j = i; j <= n; j++) {
      if (a[j] == 1)
        l++;
      else {
        r++;
        l--;
        if (l < 0) l = 0;
      }
      maxx = max(maxx, dp1[i - 1] + l + r + dp2[j + 1]);
    }
  }
  cout << maxx << endl;
  return 0;
}

```