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;
long long int sum[2005], rsum[2005];
long long int A[2005], N;
int main() {
  scanf("%lld", &N);
  for (int i = 1; i <= N; i++) {
    scanf("%lld", A + i);
    --A[i];
  }
  for (int i = 1; i <= N; i++) {
    sum[i] = sum[i - 1] + !A[i];
  }
  for (int i = N; i; i--) {
    rsum[i] = rsum[i + 1] + (A[i] == 1);
  }
  long long int fun[2];
  long long int ans = 0;
  for (int i = 1; i <= N; i++) {
    fun[0] = fun[1] = 0;
    for (int j = i; j <= N; j++) {
      fun[1] = fun[1] + (A[j] == 1);
      fun[0] = max(fun[0], fun[1]) + !A[j];
      ans = max(ans, sum[i - 1] + rsum[j + 1] + fun[0]);
    }
  }
  cout << ans;
  return 0;
}
