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 dp[2005][2005], l[2005], r[2005];
int main() {
  long long n;
  cin >> n;
  long long a[n + 5];
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i <= n; i++) l[i] = (a[i] == 1) ? l[i - 1] + 1 : l[i - 1];
  for (int i = n; i >= 1; i--) r[i] = (a[i] == 2) ? r[i + 1] + 1 : r[i + 1];
  long long c1, c2;
  for (int i = n; i >= 1; i--) {
    c1 = 0, c2 = 0;
    for (int j = i; j >= 1; j--) {
      if (a[j] == 1) c1++;
      if (a[j] == 2) c2 = max(c2, c1) + 1;
      dp[j][i] = c2;
    }
  }
  long long cnt = 0;
  for (int i = 1; i <= n; i++) {
    for (int j = i; j <= n; j++) {
      cnt = max(cnt, l[i - 1] + dp[i][j] + r[j + 1]);
    }
  }
  cout << max(cnt, max(l[n], r[0]));
}
