#include <bits/stdc++.h>
using namespace std;
const int N = 2005;
int n;
int a[N];
int d[N][N][3];
int rd[N][N][3];
int cnt1[N], cnt2[N];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", a + i);
  }
  for (int i = 1; i <= n; ++i) {
    cnt1[i] = cnt1[i - 1] + (a[i] == 1);
    cnt2[i] = cnt2[i - 1] + (a[i] == 2);
  }
  for (int i = 1; i <= n; ++i) {
    d[i][i][a[i]] = 1;
    for (int j = i + 1; j <= n; ++j) {
      if (a[j] == 1) {
        d[i][j][1] = d[i][j - 1][1] + 1;
        d[i][j][2] = max(d[i][j - 1][1], d[i][j - 1][2]);
      } else {
        d[i][j][1] = d[i][j - 1][1];
        d[i][j][2] = max(d[i][j - 1][1], d[i][j - 1][2]) + 1;
      }
    }
  }
  for (int i = 1; i <= n; ++i) {
    d[i][i][a[i]] = 1;
    for (int j = i + 1; j <= n; ++j) {
      if (a[j] == 1) {
        d[i][j][1] = d[i][j - 1][1] + 1;
        d[i][j][2] = max(d[i][j - 1][1], d[i][j - 1][2]);
      } else {
        d[i][j][1] = d[i][j - 1][1];
        d[i][j][2] = max(d[i][j - 1][1], d[i][j - 1][2]) + 1;
      }
    }
  }
  for (int j = n; j >= 1; --j) {
    rd[j][j][a[j]] = 1;
    for (int i = j - 1; i >= 1; --i) {
      if (a[i] == 1) {
        rd[i][j][1] = rd[i + 1][j][1] + 1;
        rd[i][j][2] = max(rd[i + 1][j][1], rd[i + 1][j][2]);
      } else {
        rd[i][j][1] = rd[i + 1][j][1];
        rd[i][j][2] = max(rd[i + 1][j][1], rd[i + 1][j][2]) + 1;
      }
    }
  }
  int ans = 0;
  for (int i = 1; i <= n; ++i) {
    for (int j = i; j <= n; ++j) {
      int cur = cnt1[j] - cnt1[i - 1];
      if (j + 1 <= n) {
        cur += d[j + 1][n][2];
      }
      cur += cnt1[i - 1];
      ans = max(ans, cur);
      cur = 0;
      cur += d[1][i - 1][1];
      cur += rd[i][j][2];
      if (j + 1 <= n) {
        cur += cnt2[n] - cnt2[j];
      }
      ans = max(ans, cur);
      cur = 0;
      cur += rd[1][i - 1][2];
      cur += cnt2[n] - cnt2[i - 1];
      ans = max(ans, cur);
    }
  }
  printf("%d\n", ans);
  return 0;
}

### Prompt

Your task is to create 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;
const int N = 2005;
int n;
int a[N];
int d[N][N][3];
int rd[N][N][3];
int cnt1[N], cnt2[N];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", a + i);
  }
  for (int i = 1; i <= n; ++i) {
    cnt1[i] = cnt1[i - 1] + (a[i] == 1);
    cnt2[i] = cnt2[i - 1] + (a[i] == 2);
  }
  for (int i = 1; i <= n; ++i) {
    d[i][i][a[i]] = 1;
    for (int j = i + 1; j <= n; ++j) {
      if (a[j] == 1) {
        d[i][j][1] = d[i][j - 1][1] + 1;
        d[i][j][2] = max(d[i][j - 1][1], d[i][j - 1][2]);
      } else {
        d[i][j][1] = d[i][j - 1][1];
        d[i][j][2] = max(d[i][j - 1][1], d[i][j - 1][2]) + 1;
      }
    }
  }
  for (int i = 1; i <= n; ++i) {
    d[i][i][a[i]] = 1;
    for (int j = i + 1; j <= n; ++j) {
      if (a[j] == 1) {
        d[i][j][1] = d[i][j - 1][1] + 1;
        d[i][j][2] = max(d[i][j - 1][1], d[i][j - 1][2]);
      } else {
        d[i][j][1] = d[i][j - 1][1];
        d[i][j][2] = max(d[i][j - 1][1], d[i][j - 1][2]) + 1;
      }
    }
  }
  for (int j = n; j >= 1; --j) {
    rd[j][j][a[j]] = 1;
    for (int i = j - 1; i >= 1; --i) {
      if (a[i] == 1) {
        rd[i][j][1] = rd[i + 1][j][1] + 1;
        rd[i][j][2] = max(rd[i + 1][j][1], rd[i + 1][j][2]);
      } else {
        rd[i][j][1] = rd[i + 1][j][1];
        rd[i][j][2] = max(rd[i + 1][j][1], rd[i + 1][j][2]) + 1;
      }
    }
  }
  int ans = 0;
  for (int i = 1; i <= n; ++i) {
    for (int j = i; j <= n; ++j) {
      int cur = cnt1[j] - cnt1[i - 1];
      if (j + 1 <= n) {
        cur += d[j + 1][n][2];
      }
      cur += cnt1[i - 1];
      ans = max(ans, cur);
      cur = 0;
      cur += d[1][i - 1][1];
      cur += rd[i][j][2];
      if (j + 1 <= n) {
        cur += cnt2[n] - cnt2[j];
      }
      ans = max(ans, cur);
      cur = 0;
      cur += rd[1][i - 1][2];
      cur += cnt2[n] - cnt2[i - 1];
      ans = max(ans, cur);
    }
  }
  printf("%d\n", ans);
  return 0;
}

```