#include <bits/stdc++.h>
using namespace std;
int lis[2003][2003];
int ps1[2003];
int ps2[2003];
int main() {
  int n;
  int v[2003];
  scanf("%d", &n);
  int qt1 = 0, qt2 = 0;
  int p1 = -1;
  for (int i = 0; i < n; i++) {
    scanf("%d", v + i);
    if (v[i] == 1) {
      if (p1 == -1) {
        p1 = i;
      }
      qt1++;
    } else {
      qt2++;
    }
  }
  int ans = max(qt1, qt2);
  if (p1 == -1) {
    printf("%d\n", ans);
    return 0;
  }
  for (int i = 0; i < n; i++) {
    if (v[i] == 1) {
      int aux = 1;
      reverse(v, v + i + 1);
      for (int j = 0; j < n; j++) {
        ps1[j] = v[j] == 1;
        ps2[j] = v[j] == 2;
        if (j) {
          ps1[j] += ps1[j - 1];
          ps2[j] += ps2[j - 1];
        }
      }
      for (int j = 1; j < n; j++) {
        aux = max(aux, ps1[j] + (ps2[n - 1] - ps2[j - 1]));
      }
      ans = max(ans, aux);
      reverse(v, v + i + 1);
    }
  }
  for (int j = 0; j < n; j++) {
    ps1[j] = v[j] == 1;
    ps2[j] = v[j] == 2;
    if (j) {
      ps1[j] += ps1[j - 1];
      ps2[j] += ps2[j - 1];
    }
  }
  for (int i = p1; i < n; i++) {
    ans = max(ans, ps1[i] + (ps2[n - 1] - (i ? ps2[i - 1] : 0)));
  }
  for (int i = n - 1; i >= 0; i--) {
    int l1 = 0;
    int l2 = 0;
    lis[i][i] = 1;
    if (v[i] == 1) l1 = 1;
    for (int j = i - 1; j >= 0; j--) {
      lis[i][j] = lis[i][j + 1];
      if (v[j] == 1) {
        l1++;
        lis[i][j] = max(lis[i][j], l1);
      }
      if (v[j] == 2) {
        lis[i][j]++;
      }
    }
  }
  for (int i = p1 + 1; i < n; i++) {
    for (int j = i + 1; j < n; j++) {
      ans = max(ans, lis[j][i] + ps1[i - 1] + (ps2[n - 1] - ps2[j]));
    }
  }
  printf("%d\n", ans);
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 lis[2003][2003];
int ps1[2003];
int ps2[2003];
int main() {
  int n;
  int v[2003];
  scanf("%d", &n);
  int qt1 = 0, qt2 = 0;
  int p1 = -1;
  for (int i = 0; i < n; i++) {
    scanf("%d", v + i);
    if (v[i] == 1) {
      if (p1 == -1) {
        p1 = i;
      }
      qt1++;
    } else {
      qt2++;
    }
  }
  int ans = max(qt1, qt2);
  if (p1 == -1) {
    printf("%d\n", ans);
    return 0;
  }
  for (int i = 0; i < n; i++) {
    if (v[i] == 1) {
      int aux = 1;
      reverse(v, v + i + 1);
      for (int j = 0; j < n; j++) {
        ps1[j] = v[j] == 1;
        ps2[j] = v[j] == 2;
        if (j) {
          ps1[j] += ps1[j - 1];
          ps2[j] += ps2[j - 1];
        }
      }
      for (int j = 1; j < n; j++) {
        aux = max(aux, ps1[j] + (ps2[n - 1] - ps2[j - 1]));
      }
      ans = max(ans, aux);
      reverse(v, v + i + 1);
    }
  }
  for (int j = 0; j < n; j++) {
    ps1[j] = v[j] == 1;
    ps2[j] = v[j] == 2;
    if (j) {
      ps1[j] += ps1[j - 1];
      ps2[j] += ps2[j - 1];
    }
  }
  for (int i = p1; i < n; i++) {
    ans = max(ans, ps1[i] + (ps2[n - 1] - (i ? ps2[i - 1] : 0)));
  }
  for (int i = n - 1; i >= 0; i--) {
    int l1 = 0;
    int l2 = 0;
    lis[i][i] = 1;
    if (v[i] == 1) l1 = 1;
    for (int j = i - 1; j >= 0; j--) {
      lis[i][j] = lis[i][j + 1];
      if (v[j] == 1) {
        l1++;
        lis[i][j] = max(lis[i][j], l1);
      }
      if (v[j] == 2) {
        lis[i][j]++;
      }
    }
  }
  for (int i = p1 + 1; i < n; i++) {
    for (int j = i + 1; j < n; j++) {
      ans = max(ans, lis[j][i] + ps1[i - 1] + (ps2[n - 1] - ps2[j]));
    }
  }
  printf("%d\n", ans);
}

```