#include <bits/stdc++.h>
using namespace std;
int a[2100], c1[2100], c2[2100], f[2100], ans, n;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    if (a[i] == 1)
      c1[i] = c1[i - 1] + 1, c2[i] = c2[i - 1];
    else
      c1[i] = c1[i - 1], c2[i] = c2[i - 1] + 1;
  }
  ans = 0;
  for (int i = 0; i <= n; i++) {
    int sum = ((1 <= i) ? (c1[i] - c1[(1) - 1]) : 0) +
              ((i + 1 <= n) ? (c2[n] - c2[(i + 1) - 1]) : 0);
    int maxl = 0, maxr = 0;
    for (int l = 1; l <= i; l++)
      if (((l <= i) ? (c2[i] - c2[(l)-1]) : 0) -
              ((l <= i) ? (c1[i] - c1[(l)-1]) : 0) >=
          maxl)
        maxl = ((l <= i) ? (c2[i] - c2[(l)-1]) : 0) -
               ((l <= i) ? (c1[i] - c1[(l)-1]) : 0);
    for (int r = i + 1; r <= n; r++)
      if (((i + 1 <= r) ? (c1[r] - c1[(i + 1) - 1]) : 0) -
              ((i + 1 <= r) ? (c2[r] - c2[(i + 1) - 1]) : 0) >=
          maxr)
        maxr = ((i + 1 <= r) ? (c1[r] - c1[(i + 1) - 1]) : 0) -
               ((i + 1 <= r) ? (c2[r] - c2[(i + 1) - 1]) : 0);
    if (ans < sum + maxl + maxr) ans = sum + maxl + maxr;
  }
  printf("%d", ans);
  return 0;
}

### Prompt

Generate a CPP solution to 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;
int a[2100], c1[2100], c2[2100], f[2100], ans, n;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    if (a[i] == 1)
      c1[i] = c1[i - 1] + 1, c2[i] = c2[i - 1];
    else
      c1[i] = c1[i - 1], c2[i] = c2[i - 1] + 1;
  }
  ans = 0;
  for (int i = 0; i <= n; i++) {
    int sum = ((1 <= i) ? (c1[i] - c1[(1) - 1]) : 0) +
              ((i + 1 <= n) ? (c2[n] - c2[(i + 1) - 1]) : 0);
    int maxl = 0, maxr = 0;
    for (int l = 1; l <= i; l++)
      if (((l <= i) ? (c2[i] - c2[(l)-1]) : 0) -
              ((l <= i) ? (c1[i] - c1[(l)-1]) : 0) >=
          maxl)
        maxl = ((l <= i) ? (c2[i] - c2[(l)-1]) : 0) -
               ((l <= i) ? (c1[i] - c1[(l)-1]) : 0);
    for (int r = i + 1; r <= n; r++)
      if (((i + 1 <= r) ? (c1[r] - c1[(i + 1) - 1]) : 0) -
              ((i + 1 <= r) ? (c2[r] - c2[(i + 1) - 1]) : 0) >=
          maxr)
        maxr = ((i + 1 <= r) ? (c1[r] - c1[(i + 1) - 1]) : 0) -
               ((i + 1 <= r) ? (c2[r] - c2[(i + 1) - 1]) : 0);
    if (ans < sum + maxl + maxr) ans = sum + maxl + maxr;
  }
  printf("%d", ans);
  return 0;
}

```