#include <bits/stdc++.h>
using namespace std;
const int N = 2020;
int n;
int v[N];
int c[N][2];
int pd[N][N][2];
int getQ(int l, int r, int id) {
  if (r < l) return 0;
  return c[r][id] - c[l - 1][id];
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", v + i);
    v[i]--;
    c[i][0] = c[i - 1][0] + (v[i] == 0);
    c[i][1] = c[i - 1][1] + (v[i] == 1);
  }
  int ans = max(c[n][0], c[n][1]);
  for (int i = 0; i <= n; ++i) {
    ans = max(ans, getQ(1, i, 0) + getQ(i + 1, n, 1));
  }
  for (int len = 0; len < n; ++len) {
    for (int l = 1; l + len <= n; ++l) {
      int r = l + len;
      if (len == 0) {
        pd[l][r][0] = (v[l] == 0);
        pd[l][r][1] = (v[l] == 1);
      } else {
        pd[l][r][0] = pd[l][r - 1][0];
        pd[l][r][1] = pd[l][r - 1][1];
        if (v[r] == 0) {
          pd[l][r][0] = max(pd[l][r][0], pd[l][r - 1][0] + 1);
          pd[l][r][0] = max(pd[l][r][0], pd[l][r - 1][1] + 1);
        } else {
          pd[l][r][1] = max(pd[l][r][1], pd[l][r - 1][1] + 1);
        }
      }
    }
  }
  for (int i = 1; i <= n; ++i) {
    for (int j = i; j <= n; ++j) {
      ans = max(ans, getQ(1, i - 1, 0) + max(pd[i][j][0], pd[i][j][1]) +
                         getQ(j + 1, n, 1));
    }
  }
  printf("%d\n", ans);
  return 0;
}

### Prompt

Please create a solution in Cpp 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;
const int N = 2020;
int n;
int v[N];
int c[N][2];
int pd[N][N][2];
int getQ(int l, int r, int id) {
  if (r < l) return 0;
  return c[r][id] - c[l - 1][id];
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", v + i);
    v[i]--;
    c[i][0] = c[i - 1][0] + (v[i] == 0);
    c[i][1] = c[i - 1][1] + (v[i] == 1);
  }
  int ans = max(c[n][0], c[n][1]);
  for (int i = 0; i <= n; ++i) {
    ans = max(ans, getQ(1, i, 0) + getQ(i + 1, n, 1));
  }
  for (int len = 0; len < n; ++len) {
    for (int l = 1; l + len <= n; ++l) {
      int r = l + len;
      if (len == 0) {
        pd[l][r][0] = (v[l] == 0);
        pd[l][r][1] = (v[l] == 1);
      } else {
        pd[l][r][0] = pd[l][r - 1][0];
        pd[l][r][1] = pd[l][r - 1][1];
        if (v[r] == 0) {
          pd[l][r][0] = max(pd[l][r][0], pd[l][r - 1][0] + 1);
          pd[l][r][0] = max(pd[l][r][0], pd[l][r - 1][1] + 1);
        } else {
          pd[l][r][1] = max(pd[l][r][1], pd[l][r - 1][1] + 1);
        }
      }
    }
  }
  for (int i = 1; i <= n; ++i) {
    for (int j = i; j <= n; ++j) {
      ans = max(ans, getQ(1, i - 1, 0) + max(pd[i][j][0], pd[i][j][1]) +
                         getQ(j + 1, n, 1));
    }
  }
  printf("%d\n", ans);
  return 0;
}

```