#include <bits/stdc++.h>
int dp[2010][2010];
int a[2010];
int b[3][2010];
int main() {
  int n;
  scanf("%d", &n);
  for (int k = 1; k <= n; k++) scanf("%d", &a[k]);
  for (int k = 1; k <= n; k++) {
    b[1][k] = b[1][k - 1];
    b[2][k] = b[2][k - 1];
    b[a[k]][k]++;
  }
  int p, q;
  p = q = 0;
  for (int k = 1; k <= n; k++) {
    if (a[k] == 1) {
      if (q == 0) {
        p++;
      } else {
        p++;
        q--;
      }
    } else {
      q++;
    }
  }
  int ans = p + q;
  for (int k = 1; k <= n; k++) {
    p = q = 0;
    for (int i = k; i <= n; i++) {
      if (a[i] == 2) {
        if (q == 0) {
          p++;
        } else {
          p++;
          q--;
        }
      } else {
        q++;
      }
      int s = p + q + (b[1][k - 1] - b[1][0]) + (b[2][n] - b[2][i]);
      if (s > ans) ans = s;
    }
  }
  printf("%d", ans);
}

### 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>
int dp[2010][2010];
int a[2010];
int b[3][2010];
int main() {
  int n;
  scanf("%d", &n);
  for (int k = 1; k <= n; k++) scanf("%d", &a[k]);
  for (int k = 1; k <= n; k++) {
    b[1][k] = b[1][k - 1];
    b[2][k] = b[2][k - 1];
    b[a[k]][k]++;
  }
  int p, q;
  p = q = 0;
  for (int k = 1; k <= n; k++) {
    if (a[k] == 1) {
      if (q == 0) {
        p++;
      } else {
        p++;
        q--;
      }
    } else {
      q++;
    }
  }
  int ans = p + q;
  for (int k = 1; k <= n; k++) {
    p = q = 0;
    for (int i = k; i <= n; i++) {
      if (a[i] == 2) {
        if (q == 0) {
          p++;
        } else {
          p++;
          q--;
        }
      } else {
        q++;
      }
      int s = p + q + (b[1][k - 1] - b[1][0]) + (b[2][n] - b[2][i]);
      if (s > ans) ans = s;
    }
  }
  printf("%d", ans);
}

```