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[2001], a1[2001], a2[2001];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  if (a[0] == 1)
    a1[0] = 1;
  else
    a1[0] = 0;
  for (int i = 1; i < n; i++) a1[i] = a1[i - 1] + (a[i] == 1);
  if (a[n - 1] == 2)
    a2[n - 1] == 1;
  else
    a2[n - 1] = 0;
  for (int i = n - 1; i >= 0; i--) a2[i] = a2[i + 1] + (a[i] == 2);
  int ans = 0;
  for (int k = 0; k <= n; k++) {
    int pre = 0, ber = 0;
    for (int i = 0; i <= k; i++) pre = max(pre, a1[i - 1] + a2[i] - a2[k]);
    for (int i = k; i < n; i++) ber = max(ber, a2[i + 1] + a1[i] - a1[k - 1]);
    ans = max(ans, pre + ber);
  }
  printf("%d\n", ans);
  return 0;
}
