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 n, a[2005], cnt1[2005], dp[2005][2], ni[2005][2], ans, dp0, dp1;
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i <= n; i++) cnt1[i] = cnt1[i - 1] + (a[i] == 1);
  for (int i = 1; i <= n; i++) {
    dp[i][0] = dp[i - 1][0] + (a[i] == 1);
    if (a[i] == 2)
      dp[i][1] = max(dp[i - 1][1] + 1, dp[i - 1][0] + 1);
    else
      dp[i][1] = dp[i - 1][1];
  }
  for (int i = 1; i <= n; i++) {
    ni[i][1] = ni[i + 1][1] + (a[i] == 2);
    if (a[i] == 1)
      ni[i][0] = max(ni[i + 1][0] + 1, ni[i + 1][1] + 1);
    else
      ni[i][0] = ni[i + 1][0];
  }
  for (int j = 1; j <= n; j++) {
    ans = max(ans, cnt1[j] + ni[j + 1][0]);
  }
  for (int j = 1; j <= n; j++) {
    ans = max(ans, dp[j - 1][1] + (n - j + 1 - (cnt1[n] - cnt1[j - 1])));
  }
  for (int i = 1; i <= n; i++) {
    dp0 = dp1 = 0;
    for (int j = i; j <= n; j++) {
      if (a[j] == 1)
        dp0 = max(dp1 + 1, dp0 + 1);
      else
        dp0 = dp0;
      dp1 = dp1 + (a[j] == 2);
      ans = max(ans, cnt1[i - 1] + dp0 + (n - j - (cnt1[n] - cnt1[j])));
    }
  }
  cout << ans;
  return 0;
}
