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 pre[2005][3];
int rlcs[2005][2005][3];
int main() {
  std::ios::sync_with_stdio(false);
  int n, i, j, a[2005];
  memset(pre, 0, sizeof(pre));
  memset(rlcs, 0, sizeof(rlcs));
  cin >> n;
  for (i = 1; i <= n; i++) cin >> a[i];
  for (i = 1; i <= n; i++) {
    if (a[i] == 1) {
      pre[i][1] = pre[i - 1][1] + 1;
      pre[i][2] = pre[i - 1][2];
    } else {
      pre[i][1] = pre[i - 1][1];
      pre[i][2] = pre[i - 1][2] + 1;
    }
  }
  for (i = n; i > 0; i--) {
    for (j = i; j > 0; j--) {
      if (i == j)
        rlcs[j][i][a[j]] = 1;
      else if (a[j] == 1) {
        rlcs[j][i][1] = rlcs[j + 1][i][1] + 1;
        rlcs[j][i][2] = rlcs[j + 1][i][2];
      } else {
        rlcs[j][i][1] = rlcs[j + 1][i][1];
        rlcs[j][i][2] = max(rlcs[j + 1][i][1], rlcs[j + 1][i][2]) + 1;
      }
    }
  }
  int ans = 0;
  for (i = 1; i <= n; i++) {
    for (j = i; j <= n; j++) {
      int s = pre[i - 1][1] + pre[n][2] - pre[j][2] +
              max(rlcs[i][j][1], rlcs[i][j][2]);
      ans = max(ans, s);
    }
  }
  cout << ans;
  return 0;
}
