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;
const int NMax = 2001;
int main() {
  int n, a[NMax], numberOf1[NMax], numberOf2[NMax], dp[2][NMax];
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  numberOf1[0] = (a[0] == 1);
  for (int i = 1; i < n; i++) {
    numberOf1[i] = numberOf1[i - 1] + (a[i] == 1);
  }
  dp[0][0] = 1;
  for (int i = 1; i < n; i++) {
    dp[0][i] = max(dp[0][i - 1] + (a[i] == 2), numberOf1[i]);
  }
  numberOf2[n - 1] = (a[n - 1] == 2);
  for (int i = n - 2; i > -1; i--) {
    numberOf2[i] = numberOf2[i + 1] + (a[i] == 2);
  }
  dp[1][n - 1] = 1;
  for (int i = n - 2; i > -1; i--) {
    dp[1][i] = max(dp[1][i + 1] + (a[i] == 1), numberOf2[i]);
  }
  int ans = dp[1][0];
  for (int i = 1; i < n; i++) {
    ans = max(ans, dp[0][i - 1] + dp[1][i]);
  }
  cout << ans;
  return 0;
}
