#include <bits/stdc++.h>
using namespace std;
int static const N = 2020;
int a[N], vr[N], max1[N], max0[N], desno[N], levo[N];
int main() {
  int br = 1, n, duz = 0, znak;
  cin >> n;
  for (int i = 0; i < n; i++) {
    int q;
    cin >> q;
    q--;
    if (duz == 0) {
      duz++;
      znak = q;
    } else {
      if (znak == q)
        duz++;
      else {
        a[br++] = duz;
        vr[br - 1] = znak;
        duz = 1;
        znak = q;
      }
    }
  }
  a[br++] = duz;
  vr[br - 1] = znak;
  for (int i = br - 2; i >= 0; i--) {
    desno[i] = desno[i + 1];
    if (vr[i + 1] == 1) desno[i] += a[i + 1];
  }
  for (int i = 2; i <= br; i++) {
    levo[i] = levo[i - 1];
    if (vr[i - 1] == 0) levo[i] += a[i - 1];
  }
  for (int i = br - 2; i >= 0; i--)
    if (vr[i + 1] == 0)
      max1[i] = max1[i + 1] + a[i + 1];
    else {
      max1[i] = max(max1[i + 1], desno[i]);
    }
  for (int i = 2; i <= br; i++)
    if (vr[i - 1] == 1)
      max0[i] = max0[i - 1] + a[i - 1];
    else {
      max0[i] = max(max0[i - 1], levo[i]);
    }
  int maxx = -1;
  for (int i = 0; i <= br; i++) {
    maxx = max(maxx, max1[i] + max0[i + 1]);
  }
  cout << maxx << endl;
  return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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 static const N = 2020;
int a[N], vr[N], max1[N], max0[N], desno[N], levo[N];
int main() {
  int br = 1, n, duz = 0, znak;
  cin >> n;
  for (int i = 0; i < n; i++) {
    int q;
    cin >> q;
    q--;
    if (duz == 0) {
      duz++;
      znak = q;
    } else {
      if (znak == q)
        duz++;
      else {
        a[br++] = duz;
        vr[br - 1] = znak;
        duz = 1;
        znak = q;
      }
    }
  }
  a[br++] = duz;
  vr[br - 1] = znak;
  for (int i = br - 2; i >= 0; i--) {
    desno[i] = desno[i + 1];
    if (vr[i + 1] == 1) desno[i] += a[i + 1];
  }
  for (int i = 2; i <= br; i++) {
    levo[i] = levo[i - 1];
    if (vr[i - 1] == 0) levo[i] += a[i - 1];
  }
  for (int i = br - 2; i >= 0; i--)
    if (vr[i + 1] == 0)
      max1[i] = max1[i + 1] + a[i + 1];
    else {
      max1[i] = max(max1[i + 1], desno[i]);
    }
  for (int i = 2; i <= br; i++)
    if (vr[i - 1] == 1)
      max0[i] = max0[i - 1] + a[i - 1];
    else {
      max0[i] = max(max0[i - 1], levo[i]);
    }
  int maxx = -1;
  for (int i = 0; i <= br; i++) {
    maxx = max(maxx, max1[i] + max0[i + 1]);
  }
  cout << maxx << endl;
  return 0;
}

```