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.
#include <bits/stdc++.h>
using namespace std;
struct Dbg {
  template <typename T>
  Dbg& operator,(const T& v) {
    cerr << v << ", ";
    return *this;
  }
} Dbg;
template <typename T>
ostream& operator<<(ostream& _o_, const vector<T>& _v_) {
  if (!_v_.empty()) {
    _o_ << '[';
    copy(_v_.begin(), _v_.end(), ostream_iterator<T>(_o_, ", "));
    _o_ << "\b\b]";
  }
  return _o_;
}
template <class C>
void mini(C& _a, C _b) {
  _a = min(_a, _b);
}
template <class C>
void maxi(C& _a, C _b) {
  _a = max(_a, _b);
}
int dp[4];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  vector<int> a(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  for (int i = 0; i < n; i++) {
    dp[3] = max(dp[2], dp[3]) + (a[i] == 2);
    dp[2] = max(dp[1], dp[2]) + (a[i] == 1);
    dp[1] = max(dp[0], dp[1]) + (a[i] == 2);
    dp[0] += (a[i] == 1);
  }
  cerr << "dp[0], dp[1], dp[2], dp[3]"
       << " = ";
  Dbg, dp[0], dp[1], dp[2], dp[3];
  cerr << '\n';
  ;
  cout << *max_element(dp, dp + 4);
  return 0;
}
