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;
const long long inf = 0x3f3f3f3f3f3f3f3f;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
void zz() { ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0); }
int32_t main() {
  zz();
  long long n;
  cin >> n;
  vector<long long> v(n + 1, 0);
  vector<long long> cnt1(n + 1, 0), cnt2(n + 1, 0);
  for (long long i = 1; i <= n; i++) {
    cin >> v[i];
  }
  for (long long i = 1; i <= n; i++) {
    cnt1[i] = cnt1[i - 1] + (v[i] == 1);
    cnt2[i] = cnt2[i - 1] + (v[i] == 2);
  }
  long long ms = 0;
  for (long long i = 0; i <= n; i++) {
    long long ans1 = 0;
    long long ans2 = 0;
    for (long long j = 0; j <= i; j++) {
      ans1 = max(ans1, cnt1[j] + cnt2[i] - cnt2[j]);
    }
    for (long long j = i; j <= n; j++) {
      ans2 = max(ans2, cnt1[j] - cnt1[i] + cnt2[n] - cnt2[j]);
    }
    ms = max(ms, ans1 + ans2);
  }
  cout << ms << endl;
}
