#include <bits/stdc++.h>
using namespace std;
int n, a[3000], s[3000], d[3000][30], ans = 1, s1[3000], s2[3000];
void init() {
  int i, j;
  for (i = 1; i <= n; i++) d[i][0] = a[i];
  for (j = 1; j < 30; j++)
    for (i = 1; i + (1 << j) <= n + 1; i++)
      d[i][j] = min(d[i][j - 1], d[i + (1 << (j - 1))][j - 1]);
}
int rmq(int L, int R) {
  int k = 0;
  while ((1 << (k + 1)) <= R - L + 1) k++;
  return min(d[L][k], d[R - (1 << k) + 1][k]);
}
int main() {
  int i, j;
  memset(s, 0, sizeof(s));
  cin >> n;
  for (i = 1; i <= n; i++) cin >> s[i];
  memset(a, 0, sizeof(a));
  memset(d, 0, sizeof(d));
  memset(s1, 0, sizeof(s1));
  memset(s2, 0, sizeof(s2));
  for (i = 1; i <= n; i++) {
    if (s[i] == 1) {
      s1[i] = s1[i - 1] + 1;
      s2[i] = s2[i - 1];
    } else {
      s1[i] = s1[i - 1];
      s2[i] = s2[i - 1] + 1;
    }
  }
  for (i = 1; i <= n; i++) a[i] = s1[i] - s2[i];
  init();
  for (i = 0; i <= n; i++) {
    ans = max(ans, s1[i] + s2[n] - s2[i]);
    ans = max(ans, s2[i] + s1[n] - s1[i]);
  }
  for (i = 0; i <= n - 2; i++) {
    for (j = i + 2; j <= n; j++) {
      ans = max(ans, a[i] + a[j] - rmq(i + 1, j - 1) + s2[n]);
    }
  }
  cout << 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 n, a[3000], s[3000], d[3000][30], ans = 1, s1[3000], s2[3000];
void init() {
  int i, j;
  for (i = 1; i <= n; i++) d[i][0] = a[i];
  for (j = 1; j < 30; j++)
    for (i = 1; i + (1 << j) <= n + 1; i++)
      d[i][j] = min(d[i][j - 1], d[i + (1 << (j - 1))][j - 1]);
}
int rmq(int L, int R) {
  int k = 0;
  while ((1 << (k + 1)) <= R - L + 1) k++;
  return min(d[L][k], d[R - (1 << k) + 1][k]);
}
int main() {
  int i, j;
  memset(s, 0, sizeof(s));
  cin >> n;
  for (i = 1; i <= n; i++) cin >> s[i];
  memset(a, 0, sizeof(a));
  memset(d, 0, sizeof(d));
  memset(s1, 0, sizeof(s1));
  memset(s2, 0, sizeof(s2));
  for (i = 1; i <= n; i++) {
    if (s[i] == 1) {
      s1[i] = s1[i - 1] + 1;
      s2[i] = s2[i - 1];
    } else {
      s1[i] = s1[i - 1];
      s2[i] = s2[i - 1] + 1;
    }
  }
  for (i = 1; i <= n; i++) a[i] = s1[i] - s2[i];
  init();
  for (i = 0; i <= n; i++) {
    ans = max(ans, s1[i] + s2[n] - s2[i]);
    ans = max(ans, s2[i] + s1[n] - s1[i]);
  }
  for (i = 0; i <= n - 2; i++) {
    for (j = i + 2; j <= n; j++) {
      ans = max(ans, a[i] + a[j] - rmq(i + 1, j - 1) + s2[n]);
    }
  }
  cout << ans;
  return 0;
}

```