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 int MAX_N = 1000 * 1000 + 19;
const int Mod = 1e9 + 7;
const int inf = 1 << 30;
int n, dp[MAX_N][4], a[MAX_N];
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < n; i++)
    for (int j = 0; j < 4; j++) {
      if (a[i] == 1) {
        if (!(j & 1)) {
          if (j)
            dp[i + 1][j] = max(dp[i][j], dp[i][j - 1]) + 1;
          else
            dp[i + 1][j] = dp[i][j] + 1;
        } else
          dp[i + 1][j] = dp[i][j];
      } else {
        if (j & 1)
          dp[i + 1][j] = max(dp[i][j], dp[i][j - 1]) + 1;
        else
          dp[i + 1][j] = dp[i][j];
      }
    }
  long long ans = 0;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < 4; j++) ans = max(ans, dp[i + 1][j] * 1ll);
  cout << ans;
  return 0;
}
