#include <bits/stdc++.h>
using std::cin;
using std::cout;
using std::endl;
int max(int a, int b) { return a > b ? a : b; }
const int maxn = 2000 + 10;
int n;
int a[maxn];
int front1[maxn], back1[maxn], front2[maxn], back2[maxn];
int rec[maxn];
int next2[maxn], next1[maxn];
int main() {
  int ans = 0;
  int cnt1 = 0, cnt2 = 0;
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < n; i++) {
    front1[i] = cnt1;
    front2[i] = cnt2;
    if (a[i] == 1)
      cnt1++;
    else
      cnt2++;
  }
  cnt1 = cnt2 = 0;
  for (int i = n - 1; i >= 0; i--) {
    back1[i] = cnt1;
    back2[i] = cnt2;
    if (a[i] == 1)
      cnt1++;
    else
      cnt2++;
  }
  for (int i = 0; i < n; i++) {
    int tem1 = 0, tem2 = 0;
    for (int l = 0; i - l >= 0; l++)
      tem1 = max(tem1, front1[i - l] + front2[i] - front2[i - l]);
    for (int r = 0; r + i < n; r++)
      tem2 = max(tem2, back2[r + i] + back1[i] - back1[i + r]);
    ans = max(ans, tem1 + tem2 + 1);
  }
  cout << ans << endl;
}

### 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 std::cin;
using std::cout;
using std::endl;
int max(int a, int b) { return a > b ? a : b; }
const int maxn = 2000 + 10;
int n;
int a[maxn];
int front1[maxn], back1[maxn], front2[maxn], back2[maxn];
int rec[maxn];
int next2[maxn], next1[maxn];
int main() {
  int ans = 0;
  int cnt1 = 0, cnt2 = 0;
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < n; i++) {
    front1[i] = cnt1;
    front2[i] = cnt2;
    if (a[i] == 1)
      cnt1++;
    else
      cnt2++;
  }
  cnt1 = cnt2 = 0;
  for (int i = n - 1; i >= 0; i--) {
    back1[i] = cnt1;
    back2[i] = cnt2;
    if (a[i] == 1)
      cnt1++;
    else
      cnt2++;
  }
  for (int i = 0; i < n; i++) {
    int tem1 = 0, tem2 = 0;
    for (int l = 0; i - l >= 0; l++)
      tem1 = max(tem1, front1[i - l] + front2[i] - front2[i - l]);
    for (int r = 0; r + i < n; r++)
      tem2 = max(tem2, back2[r + i] + back1[i] - back1[i + r]);
    ans = max(ans, tem1 + tem2 + 1);
  }
  cout << ans << endl;
}

```