#include <bits/stdc++.h>
using namespace std;
const int __ = 2005;
int a[__], dp[__];
int one[__], two[__];
int d[__];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = (1); i <= (n); i++) scanf("%d", &a[i]);
  int ans = 0;
  for (int i = (1); i <= (n); i++) {
    for (int j = (0); j <= (i - 1); j++)
      if (a[j] <= a[i]) dp[i] = max(dp[i], dp[j] + 1);
    ans = max(ans, dp[i]);
  }
  for (int i = (1); i <= (n); i++)
    if (a[i] == 1)
      one[i] = one[i - 1] + 1;
    else
      one[i] = one[i - 1];
  for (int i = (n); i >= (1); i--)
    if (a[i] == 2)
      two[i] = two[i + 1] + 1;
    else
      two[i] = two[i + 1];
  for (int i = (1); i <= (n); i++) {
    memset((d), (0), sizeof(d)), d[1] = -a[i];
    int idx = 1;
    for (int j = (i + 1); j <= (n); j++) {
      if (-a[j] >= d[idx])
        d[++idx] = -a[j];
      else {
        int k = upper_bound(d + 1, d + idx + 1, -a[j]) - d;
        d[k] = -a[j];
      }
      ans = max(ans, idx + one[i - 1] + two[j + 1]);
    }
  }
  printf("%d\n", ans);
  return 0;
}

### Prompt

In CPP, your task is to solve 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 __ = 2005;
int a[__], dp[__];
int one[__], two[__];
int d[__];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = (1); i <= (n); i++) scanf("%d", &a[i]);
  int ans = 0;
  for (int i = (1); i <= (n); i++) {
    for (int j = (0); j <= (i - 1); j++)
      if (a[j] <= a[i]) dp[i] = max(dp[i], dp[j] + 1);
    ans = max(ans, dp[i]);
  }
  for (int i = (1); i <= (n); i++)
    if (a[i] == 1)
      one[i] = one[i - 1] + 1;
    else
      one[i] = one[i - 1];
  for (int i = (n); i >= (1); i--)
    if (a[i] == 2)
      two[i] = two[i + 1] + 1;
    else
      two[i] = two[i + 1];
  for (int i = (1); i <= (n); i++) {
    memset((d), (0), sizeof(d)), d[1] = -a[i];
    int idx = 1;
    for (int j = (i + 1); j <= (n); j++) {
      if (-a[j] >= d[idx])
        d[++idx] = -a[j];
      else {
        int k = upper_bound(d + 1, d + idx + 1, -a[j]) - d;
        d[k] = -a[j];
      }
      ans = max(ans, idx + one[i - 1] + two[j + 1]);
    }
  }
  printf("%d\n", ans);
  return 0;
}

```