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>
#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
const int INF = 0x3f3f3f3f;
const long long INFF = 0x3f3f;
const double pi = acos(-1.0);
const double inf = 1e18;
const double eps = 1e-9;
const long long mod = 1e9 + 7;
const unsigned long long mxx = 1333331;
int a[2100];
int sum1[2100];
int sum2[2100];
int dp[2100][2100][2];
int main() {
  int n;
  cin >> n;
  int ans = -INF;
  memset(dp, 0, sizeof(dp));
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    sum1[i] += sum1[i - 1] + (a[i] == 1);
  }
  for (int i = n; i >= 1; i--) {
    sum2[i] += sum2[i + 1] + (a[i] == 2);
  }
  for (int i = 1; i <= n; i++) {
    for (int j = i; j <= n; j++) {
      dp[i][j][0] = dp[i][j - 1][0] + (a[j] == 2);
      dp[i][j][1] = max(dp[i][j - 1][0], dp[i][j - 1][1]) + (a[j] == 1);
      ans = max(ans, dp[i][j][0] + sum1[i - 1] + sum2[j + 1]);
      ans = max(ans, dp[i][j][1] + sum1[i - 1] + sum2[j + 1]);
    }
  }
  cout << ans << endl;
  return 0;
}
