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;
inline void splay(int &v) {
  v = 0;
  char c = 0;
  int p = 1;
  while (c < '0' || c > '9') {
    if (c == '-') p = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    v = (v << 3) + (v << 1) + c - '0';
    c = getchar();
  }
  v *= p;
}
const int N = 100010;
int n, a[N], b[N], dp[5], v[N], ans;
int main() {
  splay(n);
  for (int i = 1; i <= n; i++) {
    splay(v[i]);
  }
  for (int i = 1; i <= n; i++) {
    a[i] = a[i - 1] + (v[i] == 1);
  }
  for (int i = n; i >= 1; i--) {
    b[i] = b[i + 1] + (v[i] == 2);
  }
  for (int l = 1; l <= n; l++) {
    int now = 0;
    dp[1] = dp[2] = 0;
    for (int r = l; r <= n; r++) {
      if (v[r] == 1) {
        dp[1] = max(dp[1], dp[2]) + 1;
      } else {
        dp[2] = dp[2] + 1;
      }
      ans = max(ans, a[l - 1] + b[r + 1] + max(dp[1], dp[2]));
      if (l == 2) {
        int tr = 1;
      }
    }
  }
  cout << ans << endl;
}
