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 N = 2020;
int a[N], f[N][N][3], d[N][N][3];
int main() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; ++i) cin >> a[i];
  for (int i = 1; i <= n; ++i)
    for (int j = i; j <= n; ++j)
      if (a[j] == 1) {
        f[i][j][1] = f[i][j - 1][1] + 1;
        f[i][j][2] = f[i][j - 1][2];
        d[i][j][1] = max(d[i][j - 1][1], d[i][j - 1][2]) + 1;
        d[i][j][2] = d[i][j - 1][2];
      } else {
        f[i][j][1] = f[i][j - 1][1];
        f[i][j][2] = max(f[i][j - 1][1], f[i][j - 1][2]) + 1;
        d[i][j][1] = d[i][j - 1][1];
        d[i][j][2] = d[i][j - 1][2] + 1;
      }
  int ans = 0;
  for (int i = 1; i <= n; ++i)
    for (int j = i; j <= n; ++j) {
      ans = max(ans, f[1][i - 1][1] + f[i][j][1] + f[j + 1][n][2]);
      ans = max(ans, f[1][i - 1][1] + d[i][j][1] + f[j + 1][n][2]);
      ans = max(ans, f[1][i - 1][2] + d[i][j][2] + f[j + 1][n][2]);
    }
  cout << ans;
}
