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[2002], g[2002], f[2002];
int ans, p1, p2, s1, s2, s, sum, t;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", a + i);
  for (int i = 1; i <= n; ++i) f[i] = f[i - 1] + (a[i] == 1 ? 1 : 0);
  for (int i = n; i >= 1; --i) g[i] = g[i + 1] + (a[i] == 2 ? 1 : 0);
  for (int i = 1; i <= n; ++i) {
    sum = t = 0;
    s = -n;
    p1 = p2 = i;
    s1 = s2 = 0;
    for (int j = i; j >= 1; --j) {
      if (a[j] == 1)
        t--;
      else
        t++, sum++;
      if (s < t) s = t, p1 = j, s1 = sum;
    }
    sum = t = 0;
    s = -n;
    for (int j = i + 1; j <= n; ++j) {
      if (a[j] == 2)
        t--;
      else
        t++, sum++;
      if (s < t) s = t, p2 = j, s2 = sum;
    }
    ans = max(ans, f[p1 - 1] + s1 + g[i + 1]);
    ans = max(ans, f[i] + s2 + g[p2 + 1]);
    ans = max(ans, f[p1 - 1] + s1 + s2 + g[p2 + 1]);
  }
  printf("%d", ans);
  return 0;
}
