#include <bits/stdc++.h>
using namespace std;
int a[3000], n, ans, cnt, c[3000], d[3000], dp[3000];
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = n - 1; i >= 0; i--) {
    c[i] = c[i + 1];
    if (a[i] == 2) c[i]++;
  }
  for (int i = 0; i < n; i++) {
    if (i) d[i] = d[i - 1];
    if (a[i] == 1) d[i]++;
  }
  for (int i = 0; i < n; i++) {
    if (a[i] == 1) continue;
    int cnt2 = 0, cnt1 = 0;
    int j = i;
    while (j < n) {
      if (a[j] == 2) {
        cnt2++;
        cnt1--;
      } else
        cnt1++;
      j++;
      if (cnt1 < 0) cnt1 = 0;
      ans = max(ans, cnt1 + cnt2 + d[i] + c[j]);
    }
  }
  int ans1 = 0;
  fill(dp, dp + n, 100000000);
  for (int i = 0; i < n; i++) {
    *upper_bound(dp, dp + n, a[i]) = a[i];
  }
  ans1 = lower_bound(dp, dp + n, 100000000) - dp;
  cout << max(ans1, ans);
  return 0;
}

### 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;
int a[3000], n, ans, cnt, c[3000], d[3000], dp[3000];
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = n - 1; i >= 0; i--) {
    c[i] = c[i + 1];
    if (a[i] == 2) c[i]++;
  }
  for (int i = 0; i < n; i++) {
    if (i) d[i] = d[i - 1];
    if (a[i] == 1) d[i]++;
  }
  for (int i = 0; i < n; i++) {
    if (a[i] == 1) continue;
    int cnt2 = 0, cnt1 = 0;
    int j = i;
    while (j < n) {
      if (a[j] == 2) {
        cnt2++;
        cnt1--;
      } else
        cnt1++;
      j++;
      if (cnt1 < 0) cnt1 = 0;
      ans = max(ans, cnt1 + cnt2 + d[i] + c[j]);
    }
  }
  int ans1 = 0;
  fill(dp, dp + n, 100000000);
  for (int i = 0; i < n; i++) {
    *upper_bound(dp, dp + n, a[i]) = a[i];
  }
  ans1 = lower_bound(dp, dp + n, 100000000) - dp;
  cout << max(ans1, ans);
  return 0;
}

```