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>
const int kMaxN = 2000;
int a[kMaxN + 5], f[kMaxN + 5][5], b[] = {0, 1, 2, 1, 2};
inline int read() {
  register char ch = std::getchar();
  register int ans = 0;
  while (!std::isdigit(ch)) ch = std::getchar();
  while (std::isdigit(ch)) {
    ans = ans * 10 + ch - 48;
    ch = std::getchar();
  }
  return ans;
}
int main() {
  register int n;
  std::cin >> n;
  for (register int i = 1; i <= n; ++i) a[i] = read();
  for (register int i = 1; i <= n + 1; ++i)
    for (register int j = 1; j <= 4; ++j)
      *(std::begin(f[i]) + j) = std::max(f[i - 1][j], f[i - 1][j - 1]) +
                                static_cast<bool>(a[i] == b[j]);
  std::cout << *(std::end(f[n + 1]) - 1) << std::endl;
}
