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 gi() {
  int a;
  scanf("%d", &a);
  return a;
}
long long gli() {
  long long a;
  scanf("%I64d", &a);
  return a;
}
int s[2001];
int main() {
  int n = gi();
  s[0] = 0;
  for (int i = 0; i < n; i++) s[i + 1] = s[i] + gi() - 1;
  int t = 0;
  for (int i = 0; i <= n; i++) {
    int m1 = 0;
    for (int j = 0; j <= i; j++) {
      int a = j - s[j] + s[i] - s[j];
      m1 = (m1 > a ? m1 : a);
    }
    int m2 = 0;
    for (int j = 0; j <= n - i; j++) {
      int a = j - (s[i + j] - s[i]) + s[n] - s[i + j];
      m2 = (m2 > a ? m2 : a);
    }
    m1 += m2;
    t = (m1 > t ? m1 : t);
  }
  printf("%d\n", t);
  return 0;
}
