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;
long long int power(long long int x, long long int y, long long int p) {
  long long int res = 1;
  x = x % p;
  if (x == 0) return 0;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
long long int dp[2000 + 1][5];
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int n;
  cin >> n;
  long long int a[n + 1];
  for (long long int i = 1; i <= n; i++) cin >> a[i];
  for (long long int i = 1; i <= n; i++) {
    dp[i][1] = dp[i - 1][1];
    dp[i][3] = dp[i - 1][3];
    dp[i][2] = dp[i - 1][2];
    dp[i][4] = dp[i - 1][4];
    if (a[i] == 1) {
      dp[i][1] = max(dp[i][1], dp[i - 1][1] + 1);
      dp[i][3] = max(dp[i][3] + 1, dp[i - 1][2] + 1);
    } else {
      dp[i][2] = max(dp[i][2] + 1, dp[i - 1][1] + 1);
      dp[i][4] = max(dp[i][4] + 1, dp[i - 1][3] + 1);
    }
  }
  long long int ans = 1;
  for (long long int i = 1; i <= 4; i++) ans = max(ans, dp[n][i]);
  cout << ans;
}
