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 dp1[2005][2], dp2[2005][2], p[2005];
int main(int argc, char const *argv[]) {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", p + i);
  dp1[1][1] = dp2[n][1] = 1;
  for (int k = 2; k <= n; ++k) {
    dp1[k][0] = max(dp1[k - 1][0], dp1[k - 1][1]);
    dp1[k][1] = 1;
    for (int j = 1; j < k; ++j)
      if (p[j] <= p[k]) dp1[k][1] = max(dp1[k][1], dp1[j][1] + 1);
  }
  for (int k = n - 1; k > 0; --k) {
    dp2[k][0] = max(dp2[k + 1][0], dp2[k + 1][1]);
    dp2[k][1] = 1;
    for (int j = n; j > k; --j)
      if (p[j] >= p[k]) dp2[k][1] = max(dp2[k][1], dp2[j][1] + 1);
  }
  int ans = max(dp1[n][0], dp1[n][1]);
  for (int k = 1; k < n; k++) {
    ans =
        max(ans, max(dp1[k][0], dp1[k][1]) + max(dp2[k + 1][0], dp2[k + 1][1]));
  }
  printf("%d\n", ans);
  return 0;
}
