#include <bits/stdc++.h>
using namespace std;
inline int rit() {
  int f = 0, key = 1;
  char ch;
  do {
    ch = getchar();
    if (ch == '-') key = -1;
  } while (ch < '0' || ch > '9');
  do {
    f = f * 10 + ch - '0';
    ch = getchar();
  } while (ch >= '0' && ch <= '9');
  return f * key;
}
int n;
int arr[2000 + 5], suf[2000 + 5], pre[2000 + 5], ans, dp[2000 + 5][2000 + 5][3];
int32_t main(void) {
  n = rit();
  for (int i = 1; i <= (n); i++) {
    arr[i] = rit();
  }
  for (int i = 1; i <= (n); i++) {
    pre[i] = pre[i - 1] + (arr[i] == 1);
    suf[n - i + 1] = suf[n - i + 2] + (arr[n - i + 1] == 2);
  }
  for (int i = 1; i <= (n); i++)
    for (int j = i; j <= n; ++j) {
      dp[i][j][1] = max(dp[i][j - 1][1], dp[i][j - 1][2]) + (arr[j] == 1);
      dp[i][j][2] = dp[i][j - 1][2] + (arr[j] == 2);
      ans = max(ans, max(dp[i][j][1], dp[i][j][2]) + pre[i - 1] + suf[j + 1]);
    }
  printf("%d\n", ans);
  return 0;
}

### Prompt

Please formulate 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;
inline int rit() {
  int f = 0, key = 1;
  char ch;
  do {
    ch = getchar();
    if (ch == '-') key = -1;
  } while (ch < '0' || ch > '9');
  do {
    f = f * 10 + ch - '0';
    ch = getchar();
  } while (ch >= '0' && ch <= '9');
  return f * key;
}
int n;
int arr[2000 + 5], suf[2000 + 5], pre[2000 + 5], ans, dp[2000 + 5][2000 + 5][3];
int32_t main(void) {
  n = rit();
  for (int i = 1; i <= (n); i++) {
    arr[i] = rit();
  }
  for (int i = 1; i <= (n); i++) {
    pre[i] = pre[i - 1] + (arr[i] == 1);
    suf[n - i + 1] = suf[n - i + 2] + (arr[n - i + 1] == 2);
  }
  for (int i = 1; i <= (n); i++)
    for (int j = i; j <= n; ++j) {
      dp[i][j][1] = max(dp[i][j - 1][1], dp[i][j - 1][2]) + (arr[j] == 1);
      dp[i][j][2] = dp[i][j - 1][2] + (arr[j] == 2);
      ans = max(ans, max(dp[i][j][1], dp[i][j][2]) + pre[i - 1] + suf[j + 1]);
    }
  printf("%d\n", ans);
  return 0;
}

```