#include <bits/stdc++.h>
using namespace std;
int main() {
  int n;
  cin >> n;
  vector<int> a(n);
  for (int &e : a) cin >> e;
  for (int &e : a)
    if (e & 1)
      e = -1;
    else
      e = 1;
  vector<int> l[n], r[n];
  for (int i = 0; i < n; i++) {
    l[i].resize(n - i, a[i]);
    for (int j = 1; j + i < n; j++) l[i][j] = l[i][j - 1] + a[j + i];
  }
  for (int i = n - 1; i >= 0; i--) {
    r[i].resize(i + 1, a[i]);
    for (int j = 1; i - j >= 0; j++) r[i][j] = r[i][j - 1] + a[i - j];
  }
  int ans = 0;
  for (int i = 0; i <= n; i++) {
    int temp = 0;
    int lm = 0;
    int rm = 0;
    if (i > 0) {
      int ori = r[i - 1].back();
      ori = (i - abs(ori)) / 2;
      ori -= min(0, r[i - 1].back());
      lm = ori;
      for (int j = 0; i - 1 - j >= 0; j++) lm = max(ori + r[i - 1][j], lm);
    }
    if (i < n) {
      int ori = l[i].back();
      ori = (n - i - abs(ori)) / 2;
      ori += max(0, l[i].back());
      rm = ori;
      for (int j = 0; j + i < n; j++) rm = max(ori - l[i][j], rm);
    }
    temp = lm + rm;
    ans = max(temp, ans);
  }
  cout << ans;
  cin >> n;
}

### Prompt

Develop a solution in cpp 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 main() {
  int n;
  cin >> n;
  vector<int> a(n);
  for (int &e : a) cin >> e;
  for (int &e : a)
    if (e & 1)
      e = -1;
    else
      e = 1;
  vector<int> l[n], r[n];
  for (int i = 0; i < n; i++) {
    l[i].resize(n - i, a[i]);
    for (int j = 1; j + i < n; j++) l[i][j] = l[i][j - 1] + a[j + i];
  }
  for (int i = n - 1; i >= 0; i--) {
    r[i].resize(i + 1, a[i]);
    for (int j = 1; i - j >= 0; j++) r[i][j] = r[i][j - 1] + a[i - j];
  }
  int ans = 0;
  for (int i = 0; i <= n; i++) {
    int temp = 0;
    int lm = 0;
    int rm = 0;
    if (i > 0) {
      int ori = r[i - 1].back();
      ori = (i - abs(ori)) / 2;
      ori -= min(0, r[i - 1].back());
      lm = ori;
      for (int j = 0; i - 1 - j >= 0; j++) lm = max(ori + r[i - 1][j], lm);
    }
    if (i < n) {
      int ori = l[i].back();
      ori = (n - i - abs(ori)) / 2;
      ori += max(0, l[i].back());
      rm = ori;
      for (int j = 0; j + i < n; j++) rm = max(ori - l[i][j], rm);
    }
    temp = lm + rm;
    ans = max(temp, ans);
  }
  cout << ans;
  cin >> n;
}

```