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 n, a[2009], d[2009], r[2009], l[2009], ans;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    int t;
    scanf("%d", &t);
    a[i] = --t;
    d[i] = d[i - 1] + a[i];
  }
  for (int i = 1; i <= n; ++i)
    for (int k = i; k <= n; ++k) l[k] = max(l[k], i - d[i] + (d[k] - d[i - 1]));
  for (int i = 1; i <= n; ++i)
    for (int k = 1; k <= i; ++k) {
      r[k] = max(r[k], i - d[i] - (k - 1 - d[k - 1]) + d[n] - d[i - 1]);
      ans = max(ans, r[k] + l[k]);
    }
  printf("%d", min(ans, n));
  return 0;
}
