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 n;
int a[2020];
int pre[2020][4];
int sub[2020][4];
int ran[2020][4];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  int res = 0;
  int li = 0;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    pre[i][a[i]] = 1;
    pre[i][1] += pre[i - 1][1];
    pre[i][2] += li;
    li = max(pre[i][1], pre[i][2]);
  }
  li = 0;
  for (int i = n; i >= 1; i--) {
    sub[i][a[i]] = 1;
    sub[i][2] += sub[i + 1][2];
    sub[i][1] += li;
    li = max(sub[i][1], sub[i][2]);
  }
  for (int j = n; j >= 1; j--) {
    memset(ran, 0, sizeof(ran));
    li = 0;
    for (int i = j; i >= 1; i--) {
      int uno = pre[j][1] - pre[i - 1][1];
      int dos = sub[i][2] - sub[j + 1][2];
      res = max(res, pre[i - 1][1] + uno + sub[j + 1][1]);
      res = max(res, pre[i - 1][2] + dos + sub[j + 1][2]);
      ran[i][a[i]]++;
      ran[i][1] += ran[i + 1][1];
      ran[i][2] += li;
      li = max(ran[i][1], ran[i][2]);
      res = max(res, pre[i - 1][1] + li + sub[j + 1][2]);
    }
  }
  cout << res << endl;
  return 0;
}
