#include <bits/stdc++.h>
using namespace std;
long long min(long long a, long long b) {
  if (a < b)
    return a;
  else
    return b;
}
static long long gcd(long long a, long long b) {
  long long r;
  while (b != 0) {
    r = a % b;
    a = b;
    b = r;
  }
  return a;
}
long long lcm(long long a, long long b) {
  long long temp = gcd(a, b);
  return temp ? (a / temp * b) : 0;
}
unsigned long long PowMod(unsigned long long x, unsigned long long e,
                          unsigned long long mod) {
  unsigned long long res;
  if (e == 0) {
    res = 1;
  } else if (e == 1) {
    res = x;
  } else {
    res = PowMod(x, e / 2, mod);
    res = res * res % mod;
    if (e % 2) res = res * x % mod;
  }
  return res;
}
int a[2002], b[2002], one[2002][2002], two[2002][2002], cnttwo[2002],
    cntone[2002];
int solve() {
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  if (a[0] == 2) {
    cnttwo[0] = 1;
    cntone[0] = 0;
  } else {
    cnttwo[0] = 0;
    cntone[0] = 1;
  }
  int ans = 1;
  for (int i = 1; i < n; i++) {
    if (a[i] == 2) {
      cnttwo[i] = cnttwo[i - 1] + 1;
      cntone[i] = cntone[i - 1];
    } else {
      cnttwo[i] = cnttwo[i - 1];
      cntone[i] = cntone[i - 1] + 1;
    }
  }
  cnttwo[n] = cnttwo[n - 1];
  cntone[n] = cntone[n - 1];
  for (int i = 0; i < n; i++) {
    ans = max(ans, cntone[i] + cnttwo[n] - cnttwo[i]);
  }
  for (int i = n - 1; i >= 0; i--) {
    if (a[i] == 1) {
      one[i][i] = 1;
      two[i][i] = 0;
    } else {
      one[i][i] = 0;
      two[i][i] = 1;
    }
    for (int j = i - 1; j >= 0; j--) {
      if (a[j] == 1) {
        one[j][i] = one[j + 1][i] + 1;
        two[j][i] = two[j + 1][i];
      } else {
        one[j][i] = one[j + 1][i];
        two[j][i] = max(two[j + 1][i] + 1, one[j + 1][i] + 1);
      }
    }
  }
  for (int i = 0; i < n; i++) {
    for (int j = i + 1; j < n; j++) {
      ans = max(cntone[i] + two[i + 1][j] + cnttwo[n] - cnttwo[j], ans);
    }
  }
  for (int i = 0; i < n; i++) {
    ans = max(ans, two[0][i] + cnttwo[n] - cnttwo[i]);
  }
  cout << ans;
  return 0;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  for (int i = 1; i <= t; i++) {
    solve();
  }
  return 0;
}

### Prompt

Please formulate 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;
long long min(long long a, long long b) {
  if (a < b)
    return a;
  else
    return b;
}
static long long gcd(long long a, long long b) {
  long long r;
  while (b != 0) {
    r = a % b;
    a = b;
    b = r;
  }
  return a;
}
long long lcm(long long a, long long b) {
  long long temp = gcd(a, b);
  return temp ? (a / temp * b) : 0;
}
unsigned long long PowMod(unsigned long long x, unsigned long long e,
                          unsigned long long mod) {
  unsigned long long res;
  if (e == 0) {
    res = 1;
  } else if (e == 1) {
    res = x;
  } else {
    res = PowMod(x, e / 2, mod);
    res = res * res % mod;
    if (e % 2) res = res * x % mod;
  }
  return res;
}
int a[2002], b[2002], one[2002][2002], two[2002][2002], cnttwo[2002],
    cntone[2002];
int solve() {
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  if (a[0] == 2) {
    cnttwo[0] = 1;
    cntone[0] = 0;
  } else {
    cnttwo[0] = 0;
    cntone[0] = 1;
  }
  int ans = 1;
  for (int i = 1; i < n; i++) {
    if (a[i] == 2) {
      cnttwo[i] = cnttwo[i - 1] + 1;
      cntone[i] = cntone[i - 1];
    } else {
      cnttwo[i] = cnttwo[i - 1];
      cntone[i] = cntone[i - 1] + 1;
    }
  }
  cnttwo[n] = cnttwo[n - 1];
  cntone[n] = cntone[n - 1];
  for (int i = 0; i < n; i++) {
    ans = max(ans, cntone[i] + cnttwo[n] - cnttwo[i]);
  }
  for (int i = n - 1; i >= 0; i--) {
    if (a[i] == 1) {
      one[i][i] = 1;
      two[i][i] = 0;
    } else {
      one[i][i] = 0;
      two[i][i] = 1;
    }
    for (int j = i - 1; j >= 0; j--) {
      if (a[j] == 1) {
        one[j][i] = one[j + 1][i] + 1;
        two[j][i] = two[j + 1][i];
      } else {
        one[j][i] = one[j + 1][i];
        two[j][i] = max(two[j + 1][i] + 1, one[j + 1][i] + 1);
      }
    }
  }
  for (int i = 0; i < n; i++) {
    for (int j = i + 1; j < n; j++) {
      ans = max(cntone[i] + two[i + 1][j] + cnttwo[n] - cnttwo[j], ans);
    }
  }
  for (int i = 0; i < n; i++) {
    ans = max(ans, two[0][i] + cnttwo[n] - cnttwo[i]);
  }
  cout << ans;
  return 0;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  for (int i = 1; i <= t; i++) {
    solve();
  }
  return 0;
}

```