#include <bits/stdc++.h>
using namespace std;
int i, j, n;
int M[2001][2001] = {0};
void foo(int* a) {
  for (i = n - 1; i >= 0; i--) {
    M[i][i] = 1;
    int pichla1 = -1, pichla2 = -1, ones = 0;
    if (a[i] == 1) {
      ones++;
      pichla1 = i;
    } else {
      pichla2 = i;
    }
    for (j = i - 1; j >= 0; j--) {
      if (a[j] == 1) {
        ones++;
        M[j][i] = max(ones, M[j + 1][i]);
        pichla1 = j;
      } else {
        int ali = 0, bli = 0;
        ali = ones + 1;
        if (pichla2 != -1) {
          bli = M[pichla2][i] + 1;
        }
        M[j][i] = max(max(ali, bli), M[j + 1][i]);
        pichla2 = j;
      }
    }
  }
}
int main() {
  std::cout.precision(0);
  ios_base::sync_with_stdio(false);
  cout << fixed;
  cin >> n;
  int a[n];
  for (i = 0; i < n; i++) cin >> a[i];
  int cash1[2001] = {0}, cash2[2001] = {0};
  if (a[0] == 1)
    cash1[0]++;
  else
    cash2[0]++;
  for (i = 1; i < n; i++) {
    cash1[i] = cash1[i - 1];
    cash2[i] = cash2[i - 1];
    if (a[i] == 1)
      cash1[i]++;
    else
      cash2[i]++;
  }
  foo(a);
  int res = 0;
  for (i = 0; i < n; i++) {
    for (j = i; j < n; j++) {
      int left = 0, right = 0;
      if (i != 0) left = cash1[i - 1];
      if (j != 0) right = cash2[n - 1] - cash2[j];
      int ans = left + M[i][j] + right;
      res = max(ans, res);
    }
  }
  cout << res;
}

### Prompt

Your challenge is to write 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 i, j, n;
int M[2001][2001] = {0};
void foo(int* a) {
  for (i = n - 1; i >= 0; i--) {
    M[i][i] = 1;
    int pichla1 = -1, pichla2 = -1, ones = 0;
    if (a[i] == 1) {
      ones++;
      pichla1 = i;
    } else {
      pichla2 = i;
    }
    for (j = i - 1; j >= 0; j--) {
      if (a[j] == 1) {
        ones++;
        M[j][i] = max(ones, M[j + 1][i]);
        pichla1 = j;
      } else {
        int ali = 0, bli = 0;
        ali = ones + 1;
        if (pichla2 != -1) {
          bli = M[pichla2][i] + 1;
        }
        M[j][i] = max(max(ali, bli), M[j + 1][i]);
        pichla2 = j;
      }
    }
  }
}
int main() {
  std::cout.precision(0);
  ios_base::sync_with_stdio(false);
  cout << fixed;
  cin >> n;
  int a[n];
  for (i = 0; i < n; i++) cin >> a[i];
  int cash1[2001] = {0}, cash2[2001] = {0};
  if (a[0] == 1)
    cash1[0]++;
  else
    cash2[0]++;
  for (i = 1; i < n; i++) {
    cash1[i] = cash1[i - 1];
    cash2[i] = cash2[i - 1];
    if (a[i] == 1)
      cash1[i]++;
    else
      cash2[i]++;
  }
  foo(a);
  int res = 0;
  for (i = 0; i < n; i++) {
    for (j = i; j < n; j++) {
      int left = 0, right = 0;
      if (i != 0) left = cash1[i - 1];
      if (j != 0) right = cash2[n - 1] - cash2[j];
      int ans = left + M[i][j] + right;
      res = max(ans, res);
    }
  }
  cout << res;
}

```