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;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  vector<int> a(n + 1), pr1(n + 1, 0), pr2(n + 1, 0);
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    pr1[i] = pr1[i - 1] + (a[i] == 1);
    pr2[i] = pr2[i - 1] + (a[i] == 2);
  }
  vector<int> ik(n + 1, 0), jk(n + 1, 0);
  for (int i = 1; i <= n; i++) {
    for (int k = i; k <= n; k++) {
      int f = pr2[k] - pr2[i - 1] + pr1[i];
      ik[k] = max(ik[k], f);
    }
  }
  for (int j = n; j > 0; j--) {
    for (int k = j; k > 0; k--) {
      int f = pr1[j] - pr1[k - 1] + pr2[n] - pr2[j - 1];
      jk[k] = max(jk[k], f);
    }
  }
  int mx = 0;
  for (int k = 1; k <= n; k++) {
    mx = max(mx, ik[k] + jk[k]);
  }
  cout << min(mx, n);
}
