#include <bits/stdc++.h>
using namespace std;
const int MOD = 1e9 + 7;
const int N = 1e6 + 5;
const int K = (1 << 17) + 10;
const long long int INF = 2e18 + 500;
const int inf = 0x7fffffff;
const int SQRT = 500;
const int lgN = 20;
int T[N], O[N];
int st[N][4];
int n;
int dp[2003][2003][3];
int A[N], B[N];
int ones(int l, int r) {
  if (l < 1 || r > n) return 0;
  return O[r] - O[l - 1];
}
int twos(int l, int r) {
  if (l < 1 || r > n) return 0;
  return T[r] - T[l - 1];
}
int flip(int l, int r) { return max(dp[l][r][1], dp[l][r][2]); }
int main(int argc, char const *argv[]) {
  cin.sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> A[i];
    B[i] = A[i];
    O[i] = O[i - 1] + (A[i] == 1);
    T[i] = T[i - 1] + (A[i] == 2);
  }
  for (int i = n; i >= 1; i--) {
    for (int j = i; j >= 1; j--) {
      dp[j][i][1] = dp[j + 1][i][1];
      dp[j][i][2] = dp[j + 1][i][2];
      if (A[j] == 1) {
        dp[j][i][1] = 1 + dp[j + 1][i][1];
      } else {
        dp[j][i][2] = 1 + max(dp[j + 1][i][1], dp[j + 1][i][2]);
      }
    }
  }
  int ans = 0;
  for (int l = 1; l <= n; l++) {
    for (int r = l; r <= n; r++) {
      ans = max(ans, ones(1, l - 1) + ones(l, r) +
                         max(ones(r + 1, n), twos(r + 1, n)));
      ans = max(ans, max(ones(1, l - 1), twos(1, l - 1)) + twos(l, r) +
                         twos(r + 1, n));
      ans = max(ans, ones(1, l - 1) + flip(l, r) + twos(r + 1, n));
    }
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int MOD = 1e9 + 7;
const int N = 1e6 + 5;
const int K = (1 << 17) + 10;
const long long int INF = 2e18 + 500;
const int inf = 0x7fffffff;
const int SQRT = 500;
const int lgN = 20;
int T[N], O[N];
int st[N][4];
int n;
int dp[2003][2003][3];
int A[N], B[N];
int ones(int l, int r) {
  if (l < 1 || r > n) return 0;
  return O[r] - O[l - 1];
}
int twos(int l, int r) {
  if (l < 1 || r > n) return 0;
  return T[r] - T[l - 1];
}
int flip(int l, int r) { return max(dp[l][r][1], dp[l][r][2]); }
int main(int argc, char const *argv[]) {
  cin.sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> A[i];
    B[i] = A[i];
    O[i] = O[i - 1] + (A[i] == 1);
    T[i] = T[i - 1] + (A[i] == 2);
  }
  for (int i = n; i >= 1; i--) {
    for (int j = i; j >= 1; j--) {
      dp[j][i][1] = dp[j + 1][i][1];
      dp[j][i][2] = dp[j + 1][i][2];
      if (A[j] == 1) {
        dp[j][i][1] = 1 + dp[j + 1][i][1];
      } else {
        dp[j][i][2] = 1 + max(dp[j + 1][i][1], dp[j + 1][i][2]);
      }
    }
  }
  int ans = 0;
  for (int l = 1; l <= n; l++) {
    for (int r = l; r <= n; r++) {
      ans = max(ans, ones(1, l - 1) + ones(l, r) +
                         max(ones(r + 1, n), twos(r + 1, n)));
      ans = max(ans, max(ones(1, l - 1), twos(1, l - 1)) + twos(l, r) +
                         twos(r + 1, n));
      ans = max(ans, ones(1, l - 1) + flip(l, r) + twos(r + 1, n));
    }
  }
  cout << ans << endl;
  return 0;
}

```