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;
const int maxn = 2e3 + 20;
int p12[maxn], s12[maxn], t1[maxn], t2[maxn];
int a[maxn];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i <= n; i++) {
    t1[i] = t1[i - 1] + (a[i] == 1);
    t2[i] = t2[i - 1] + (a[i] == 2);
    p12[i] = max(p12[i - 1] + (a[i] == 2), max(t1[i], t2[i]));
  }
  for (int i = n; i >= 1; i--)
    s12[i] = max(s12[i + 1] + (a[i] == 1),
                 max(t1[n] - t1[i - 1], t2[n] - t2[i - 1]));
  int res = 0;
  for (int l = 1; l <= n; l++) {
    int x12 = 0, x1 = 0, x2 = 0;
    for (int r = l; r <= n; r++) {
      x1 += (a[r] == 1);
      x2 += (a[r] == 2);
      x12 = max(x12 + (a[r] == 1), max(x1, x2));
      res = max(res, t1[l - 1] + x12 + (t2[n] - t2[r]));
      res = max(res, p12[l - 1] + x2 + (t2[n] - t2[r]));
      res = max(res, t1[l - 1] + x1 + s12[r + 1]);
    }
  }
  cout << res << endl;
}
