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 a[3000], n;
int h1[3000], h2[3000], d1[3000], d2[3000], ans;
int main(int argc, const char* argv[]) {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    h1[i] = h1[i - 1];
    h2[i] = h2[i - 1];
    if (a[i] == 2) h2[i]++;
    if (a[i] == 1) h1[i]++;
    h2[i] = max(h2[i], h1[i]);
  }
  for (int i = n; i >= 1; i--) {
    d1[i] = d1[i + 1];
    d2[i] = d2[i + 1];
    if (a[i] == 1)
      d1[i]++;
    else
      d2[i]++;
    d1[i] = max(d1[i], d2[i]);
    ans = max(ans, d1[i] + h2[i - 1]);
  }
  cout << ans << endl;
}
