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[2005], pre1[2005], aft2[2005];
int n;
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    pre1[i] = pre1[i - 1] + (a[i] == 1);
  }
  for (int i = n; i >= 1; i--) aft2[i] = aft2[i + 1] + (a[i] == 2);
  int ans = 0;
  int kk, s1, s2;
  for (int k = 1; k <= n; k++) {
    int sum1 = 0, sum2 = 0;
    for (int i = 1; i <= k; i++)
      sum1 = max(sum1, pre1[i] + aft2[i] - aft2[k + 1]);
    for (int i = k; i <= n; i++)
      sum2 = max(sum2, aft2[i + 1] + pre1[i] - pre1[k]);
    if (ans < sum1 + sum2) {
      ans = sum1 + sum2;
      kk = k;
      s1 = sum1;
      s2 = sum2;
    }
  }
  cout << ans << endl;
  return 0;
}
