#include <bits/stdc++.h>
using namespace std;
int d1[2005], d2[2005], d3[2005], g[2005], a[2005], n, ans;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1, t = 0; i <= n; i++) {
    int k = upper_bound(g + 1, g + t + 1, a[i]) - g;
    t = max(d1[i] = k, t);
    g[k] = a[i];
  }
  for (int i = n, t = 0; i > 0; i--) {
    int k = upper_bound(g + 1, g + t + 1, a[i], greater<int>()) - g;
    t = max(d2[i] = k, t);
    g[k] = a[i];
  }
  for (int i = 1; i <= n; i++) {
    for (int j = i, t = 0; j <= n; j++) {
      int k = upper_bound(g + 1, g + t + 1, a[j], greater<int>()) - g;
      t = max(d3[j] = k, t);
      g[k] = a[j];
      ans = max(ans, d1[i - 1] + d2[j] + d3[j] - 1);
    }
    ans = max(ans, d1[i - 1] + *max_element(d3 + i, d3 + n + 1) - 1);
  }
  printf("%d", ans);
  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;
int d1[2005], d2[2005], d3[2005], g[2005], a[2005], n, ans;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1, t = 0; i <= n; i++) {
    int k = upper_bound(g + 1, g + t + 1, a[i]) - g;
    t = max(d1[i] = k, t);
    g[k] = a[i];
  }
  for (int i = n, t = 0; i > 0; i--) {
    int k = upper_bound(g + 1, g + t + 1, a[i], greater<int>()) - g;
    t = max(d2[i] = k, t);
    g[k] = a[i];
  }
  for (int i = 1; i <= n; i++) {
    for (int j = i, t = 0; j <= n; j++) {
      int k = upper_bound(g + 1, g + t + 1, a[j], greater<int>()) - g;
      t = max(d3[j] = k, t);
      g[k] = a[j];
      ans = max(ans, d1[i - 1] + d2[j] + d3[j] - 1);
    }
    ans = max(ans, d1[i - 1] + *max_element(d3 + i, d3 + n + 1) - 1);
  }
  printf("%d", ans);
  return 0;
}

```