#include <bits/stdc++.h>
using namespace std;
const int inf = (int)1e9;
const long long linf = (long long)1e18;
const int mod = (int)1e9 + 7;
const long double eps = (long double)1e-8;
const int maxn = (int)1e5 + 5;
const long double pi = acos(-1);
int n, a[maxn], sufs2[maxn];
int d[maxn];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (auto i = 0; i < n; ++i) {
    cin >> a[i];
  }
  for (auto i = 0; i < n; ++i) {
    int s[3];
    memset((s), 0, sizeof(s));
    d[i] = 0;
    for (auto j = i - 1; j >= 0; --j) {
      ++s[a[j]];
      d[i] = max(d[i], s[2] - s[1]);
    }
    d[i] += s[1];
  }
  memset((sufs2), 0, sizeof(sufs2));
  for (auto i = n - 1; i >= 0; --i) {
    sufs2[i] = sufs2[i + 1] + (a[i] == 2);
  }
  int ans = 0;
  for (auto i = 0; i < n; ++i) ans += (a[i] == 2);
  for (auto i = 0; i < n; ++i) {
    int s[3];
    memset((s), 0, sizeof(s));
    for (auto j = i; j < n; ++j) {
      ++s[a[j]];
      ans = max(ans, d[i] + s[1] + sufs2[j + 1]);
    }
  }
  cout << ans << endl;
  return 0;
}

### Prompt

In CPP, your task is to solve 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 inf = (int)1e9;
const long long linf = (long long)1e18;
const int mod = (int)1e9 + 7;
const long double eps = (long double)1e-8;
const int maxn = (int)1e5 + 5;
const long double pi = acos(-1);
int n, a[maxn], sufs2[maxn];
int d[maxn];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (auto i = 0; i < n; ++i) {
    cin >> a[i];
  }
  for (auto i = 0; i < n; ++i) {
    int s[3];
    memset((s), 0, sizeof(s));
    d[i] = 0;
    for (auto j = i - 1; j >= 0; --j) {
      ++s[a[j]];
      d[i] = max(d[i], s[2] - s[1]);
    }
    d[i] += s[1];
  }
  memset((sufs2), 0, sizeof(sufs2));
  for (auto i = n - 1; i >= 0; --i) {
    sufs2[i] = sufs2[i + 1] + (a[i] == 2);
  }
  int ans = 0;
  for (auto i = 0; i < n; ++i) ans += (a[i] == 2);
  for (auto i = 0; i < n; ++i) {
    int s[3];
    memset((s), 0, sizeof(s));
    for (auto j = i; j < n; ++j) {
      ++s[a[j]];
      ans = max(ans, d[i] + s[1] + sufs2[j + 1]);
    }
  }
  cout << ans << endl;
  return 0;
}

```