#include <bits/stdc++.h>
using namespace std;
const int MOD = 1000000007;
const long long int LLINF = 1ll << 60;
long long int sti(string a) {
  long long int k = 0;
  for (int i = 0; i < a.size(); ++i) {
    k *= 10;
    k += a[i] - '0';
  }
  return k;
}
string its(long long int k) {
  string a;
  while (k) {
    a.push_back((k % 10) + '0');
    k /= 10;
  }
  reverse(a.begin(), a.end());
  return a;
}
int gcd(int a, int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
template <class T, class K>
pair<T, K> mp(T a, K b) {
  return make_pair(a, b);
}
long long int mult(long long int a, long long int b) { return (a * b) % MOD; }
long long int pot(long long int n, long long int k) {
  if (k == 0) return 1;
  long long int a = pot(n, k / 2);
  a = mult(a, a);
  if (k % 2)
    return mult(a, n);
  else
    return a;
}
long long int divide(long long int a, long long int b) {
  return mult(a, pot(b, MOD - 2));
}
long long int sub(long long int a, long long int b) {
  return (a - b + MOD) % MOD;
}
long long int add(long long int a, long long int b) { return (a + b) % MOD; }
bool prime(long long int a) {
  if (a == 1) return 0;
  for (int i = 2; i <= round(sqrt(a)); ++i) {
    if (a % i == 0) return 0;
  }
  return 1;
}
int n;
int sus[2001][2001];
int h[2001], be1[2001], af2[2001], ma;
int main() {
  std::ios_base::sync_with_stdio(false);
  cin >> n;
  for (int i = 0; i < n; ++i) {
    cin >> h[i];
    if (h[i] == 1) ++be1[i];
    if (i) be1[i] += be1[i - 1];
  }
  for (int i = n - 1; i >= 0; --i) {
    if (h[i] == 2) ++af2[i];
    af2[i] += af2[i + 1];
    int ma12 = -MOD, n2 = 0, n1 = 0;
    for (int j = i; j >= 0; --j) {
      if (h[j] == 2)
        ++n2;
      else
        ++n1;
      ma12 = max(n1 - n2, ma12);
      sus[j][i] = ma12 + n2;
      if (!n1) ++sus[j][i];
    }
  }
  for (int i = 0; i < n; ++i) {
    for (int j = i; j < n; ++j) {
      ma = max(ma, sus[i][j] + (i ? be1[i - 1] : 0) +
                       ((j < n - 1) ? af2[j + 1] : 0));
    }
  }
  cout << ma << endl;
}

### Prompt

Your challenge is to write a CPP solution to the following problem:

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 = 1000000007;
const long long int LLINF = 1ll << 60;
long long int sti(string a) {
  long long int k = 0;
  for (int i = 0; i < a.size(); ++i) {
    k *= 10;
    k += a[i] - '0';
  }
  return k;
}
string its(long long int k) {
  string a;
  while (k) {
    a.push_back((k % 10) + '0');
    k /= 10;
  }
  reverse(a.begin(), a.end());
  return a;
}
int gcd(int a, int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
template <class T, class K>
pair<T, K> mp(T a, K b) {
  return make_pair(a, b);
}
long long int mult(long long int a, long long int b) { return (a * b) % MOD; }
long long int pot(long long int n, long long int k) {
  if (k == 0) return 1;
  long long int a = pot(n, k / 2);
  a = mult(a, a);
  if (k % 2)
    return mult(a, n);
  else
    return a;
}
long long int divide(long long int a, long long int b) {
  return mult(a, pot(b, MOD - 2));
}
long long int sub(long long int a, long long int b) {
  return (a - b + MOD) % MOD;
}
long long int add(long long int a, long long int b) { return (a + b) % MOD; }
bool prime(long long int a) {
  if (a == 1) return 0;
  for (int i = 2; i <= round(sqrt(a)); ++i) {
    if (a % i == 0) return 0;
  }
  return 1;
}
int n;
int sus[2001][2001];
int h[2001], be1[2001], af2[2001], ma;
int main() {
  std::ios_base::sync_with_stdio(false);
  cin >> n;
  for (int i = 0; i < n; ++i) {
    cin >> h[i];
    if (h[i] == 1) ++be1[i];
    if (i) be1[i] += be1[i - 1];
  }
  for (int i = n - 1; i >= 0; --i) {
    if (h[i] == 2) ++af2[i];
    af2[i] += af2[i + 1];
    int ma12 = -MOD, n2 = 0, n1 = 0;
    for (int j = i; j >= 0; --j) {
      if (h[j] == 2)
        ++n2;
      else
        ++n1;
      ma12 = max(n1 - n2, ma12);
      sus[j][i] = ma12 + n2;
      if (!n1) ++sus[j][i];
    }
  }
  for (int i = 0; i < n; ++i) {
    for (int j = i; j < n; ++j) {
      ma = max(ma, sus[i][j] + (i ? be1[i - 1] : 0) +
                       ((j < n - 1) ? af2[j + 1] : 0));
    }
  }
  cout << ma << endl;
}

```