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 a[2005];
int dp1[2005][2005], dp2[2005][2005], dp3[2005][2005], dp4[2005][2005];
void get_dp(int n) {
  for (int i = 1; i <= n; ++i) {
    for (int j = i; j <= n; ++j) {
      if (i == j) {
        dp1[i][j] = (a[j] == 1);
        continue;
      }
      dp1[i][j] = dp1[i][j - 1] + (a[j] == 1);
    }
  }
  for (int i = 1; i <= n; ++i) {
    for (int j = i; j <= n; ++j) {
      if (i == j) {
        dp2[i][j] = (a[j] == 2);
        continue;
      }
      dp2[i][j] = dp2[i][j - 1] + (a[j] == 2);
    }
  }
  for (int i = 1; i <= n; ++i) {
    for (int j = i + 1; j <= n; ++j) {
      dp3[i][j] = max(dp3[i][j], dp1[i][j - 1] + dp2[j][j]);
      dp3[i][j] = max(dp3[i][j], dp3[i][j - 1] + dp2[j][j]);
      dp3[i][j] = max(dp3[i][j], dp1[i][j - 1] + dp3[j][j]);
    }
  }
  for (int i = 1; i <= n; ++i) {
    for (int j = i + 1; j <= n; ++j) {
      dp4[i][j] = max(dp4[i][j], dp2[i][j - 1] + dp1[j][j]);
      dp4[i][j] = max(dp4[i][j], dp4[i][j - 1] + dp1[j][j]);
      dp3[i][j] = max(dp4[i][j], dp2[i][j - 1] + dp4[j][j]);
    }
  }
}
int main() {
  memset(a, 0, sizeof(a));
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  get_dp(n);
  int ans = 0;
  for (int i = 1; i <= n; ++i) {
    for (int j = i; j <= n; ++j) {
      ans = max(ans, dp1[1][i - 1] + dp1[i][j] + dp1[j + 1][n]);
      ans = max(ans, dp1[1][i - 1] + dp1[i][j] + dp2[j + 1][n]);
      ans = max(ans, dp1[1][i - 1] + dp1[i][j] + dp3[j + 1][n]);
      ans = max(ans, dp1[1][i - 1] + dp2[i][j] + dp2[j + 1][n]);
      ans = max(ans, dp1[1][i - 1] + dp4[i][j] + dp2[j + 1][n]);
      ans = max(ans, dp2[1][i - 1] + dp2[i][j] + dp2[j + 1][n]);
      ans = max(ans, dp3[1][i - 1] + dp2[i][j] + dp2[j + 1][n]);
    }
  }
  cout << ans << endl;
  return 0;
}
