#include <bits/stdc++.h>
using namespace std;
const int MAXN = 2e3 + 23;
int n, a[MAXN];
int jen[MAXN][MAXN], dva[MAXN][MAXN];
int dpl[MAXN], dpr[MAXN];
int main() {
  cin.sync_with_stdio(false);
  cin >> n;
  int ones = 0;
  int twos = 0;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    if (a[i] == 1) {
      ones++;
    } else {
      twos++;
    }
  }
  int best = max(ones, twos);
  for (int i = 0; i < n; i++) {
    if (a[i] == 1) {
      jen[i][i] = 1;
    } else {
      dva[i][i] = 1;
    }
    for (int j = i + 1; j < n; j++) {
      jen[i][j] += jen[i][j - 1];
      dva[i][j] += dva[i][j - 1];
      if (a[j] == 1) {
        jen[i][j]++;
      } else {
        dva[i][j]++;
      }
    }
  }
  for (int i = 0; i < n; i++) {
    dpr[i] = max(jen[i][n - 1], dva[i][n - 1]);
    for (int j = i; j < n; j++) {
      dpr[i] = max(dpr[i], jen[i][j] + dva[j + 1][n - 1]);
    }
  }
  for (int i = 0; i < n; i++) {
    dpl[i] = max(jen[0][i], dva[0][i]);
    for (int j = 0; j <= i; j++) {
      dpl[i] = max(dpl[i], jen[0][j] + dva[j + 1][i]);
    }
  }
  assert(dpl[n - 1] == dpr[0]);
  for (int i = 0; i < n; i++) {
    best = max(best, dpl[i] + dpr[i + 1]);
  }
  cout << best << endl;
  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 MAXN = 2e3 + 23;
int n, a[MAXN];
int jen[MAXN][MAXN], dva[MAXN][MAXN];
int dpl[MAXN], dpr[MAXN];
int main() {
  cin.sync_with_stdio(false);
  cin >> n;
  int ones = 0;
  int twos = 0;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    if (a[i] == 1) {
      ones++;
    } else {
      twos++;
    }
  }
  int best = max(ones, twos);
  for (int i = 0; i < n; i++) {
    if (a[i] == 1) {
      jen[i][i] = 1;
    } else {
      dva[i][i] = 1;
    }
    for (int j = i + 1; j < n; j++) {
      jen[i][j] += jen[i][j - 1];
      dva[i][j] += dva[i][j - 1];
      if (a[j] == 1) {
        jen[i][j]++;
      } else {
        dva[i][j]++;
      }
    }
  }
  for (int i = 0; i < n; i++) {
    dpr[i] = max(jen[i][n - 1], dva[i][n - 1]);
    for (int j = i; j < n; j++) {
      dpr[i] = max(dpr[i], jen[i][j] + dva[j + 1][n - 1]);
    }
  }
  for (int i = 0; i < n; i++) {
    dpl[i] = max(jen[0][i], dva[0][i]);
    for (int j = 0; j <= i; j++) {
      dpl[i] = max(dpl[i], jen[0][j] + dva[j + 1][i]);
    }
  }
  assert(dpl[n - 1] == dpr[0]);
  for (int i = 0; i < n; i++) {
    best = max(best, dpl[i] + dpr[i + 1]);
  }
  cout << best << endl;
  return 0;
}

```