#include <bits/stdc++.h>
using namespace std;
int a[2005], sm1[2005], sm2[2005], b[2005];
int dp[2005][2005], f[2005][2];
int main(void) {
  int n, ans = 0;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]), b[i] = a[i];
  for (int i = n; i > 0; i--) {
    memset(f, 0, sizeof(f));
    dp[i][i] = 1;
    if (b[i] == 1)
      f[i][1] = f[i][2] = 1;
    else
      f[i][2] = 1;
    for (int j = i - 1; j > 0; j--) {
      if (b[j] == 1) {
        f[j][1] = f[j + 1][1] + 1;
        f[j][2] = max(f[j + 1][2], f[j][1]);
      } else {
        f[j][1] = f[j + 1][1];
        f[j][2] = max(f[j][1], f[j + 1][2] + 1);
      }
      dp[j][i] = f[j][2];
    }
  }
  for (int i = 1; i <= n; i++) {
    sm1[i] = sm1[i - 1] + (a[i] == 2);
    sm2[i] = sm2[i - 1] + (a[i] == 1);
  }
  ans = max(sm1[n], sm2[n]);
  for (int i = 1; i <= n; i++)
    for (int j = i; j <= n; j++) {
      ans = max(ans, sm2[i - 1] + sm1[n] - sm1[j] + dp[i][j]);
    }
  printf("%d\n", 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 a[2005], sm1[2005], sm2[2005], b[2005];
int dp[2005][2005], f[2005][2];
int main(void) {
  int n, ans = 0;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]), b[i] = a[i];
  for (int i = n; i > 0; i--) {
    memset(f, 0, sizeof(f));
    dp[i][i] = 1;
    if (b[i] == 1)
      f[i][1] = f[i][2] = 1;
    else
      f[i][2] = 1;
    for (int j = i - 1; j > 0; j--) {
      if (b[j] == 1) {
        f[j][1] = f[j + 1][1] + 1;
        f[j][2] = max(f[j + 1][2], f[j][1]);
      } else {
        f[j][1] = f[j + 1][1];
        f[j][2] = max(f[j][1], f[j + 1][2] + 1);
      }
      dp[j][i] = f[j][2];
    }
  }
  for (int i = 1; i <= n; i++) {
    sm1[i] = sm1[i - 1] + (a[i] == 2);
    sm2[i] = sm2[i - 1] + (a[i] == 1);
  }
  ans = max(sm1[n], sm2[n]);
  for (int i = 1; i <= n; i++)
    for (int j = i; j <= n; j++) {
      ans = max(ans, sm2[i - 1] + sm1[n] - sm1[j] + dp[i][j]);
    }
  printf("%d\n", ans);
  return 0;
}

```