#include <bits/stdc++.h>
using namespace std;
const long long infl = 0x3f3f3f3f3f3f3f3fLL;
const long long infi = 0x3f3f3f3f;
const long long mod = infl;
long long mygcd(long long a, long long b) {
  if (b == 0) {
    return a;
  }
  return mygcd(b, a % b);
}
long long mypow(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = (res * a) % mod;
    a = (a * a) % mod;
    b >>= 1;
  }
  return res;
}
void solve() {
  long long n;
  cin >> n;
  vector<long long> inp(n), pre(n), suf(n);
  for (long long i = 0; i < n; i++) {
    cin >> inp[i];
  }
  if (inp[0] == 1) {
    pre[0] = 1;
  }
  for (long long i = 1; i < n; i++) {
    if (inp[i] == 1) {
      pre[i] = pre[i - 1] + 1;
    } else {
      pre[i] = pre[i - 1];
    }
  }
  if (inp[n - 1] == 2) {
    suf[n - 1] = 1;
  }
  for (long long i = n - 2; i >= 0; i--) {
    if (inp[i] == 2) {
      suf[i] = suf[i + 1] + 1;
    } else {
      suf[i] = suf[i + 1];
    }
  }
  long long finalans = 0;
  for (long long i = 0; i < n; i++) {
    long long p = 0;
    if (i > 0) {
      p = pre[i - 1];
    }
    long long a1 = p;
    long long a2 = p;
    for (long long j = i; j < n; j++) {
      if (inp[j] == 1) {
        a1 = max(a1, a2) + 1;
      } else {
        a2++;
      }
      long long tempans = max(a1, a2);
      long long s = 0;
      if (j != n - 1) {
        s = suf[j + 1];
      }
      tempans += s;
      finalans = max(finalans, tempans);
    }
  }
  cout << finalans << '\n';
}
signed main() {
  ios_base::sync_with_stdio(false);
  cout << fixed << setprecision(20);
  cin.tie(NULL);
  long long t = 1;
  while (t--) {
    solve();
  }
  return 0;
}

### Prompt

Please create a solution in CPP 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 long long infl = 0x3f3f3f3f3f3f3f3fLL;
const long long infi = 0x3f3f3f3f;
const long long mod = infl;
long long mygcd(long long a, long long b) {
  if (b == 0) {
    return a;
  }
  return mygcd(b, a % b);
}
long long mypow(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = (res * a) % mod;
    a = (a * a) % mod;
    b >>= 1;
  }
  return res;
}
void solve() {
  long long n;
  cin >> n;
  vector<long long> inp(n), pre(n), suf(n);
  for (long long i = 0; i < n; i++) {
    cin >> inp[i];
  }
  if (inp[0] == 1) {
    pre[0] = 1;
  }
  for (long long i = 1; i < n; i++) {
    if (inp[i] == 1) {
      pre[i] = pre[i - 1] + 1;
    } else {
      pre[i] = pre[i - 1];
    }
  }
  if (inp[n - 1] == 2) {
    suf[n - 1] = 1;
  }
  for (long long i = n - 2; i >= 0; i--) {
    if (inp[i] == 2) {
      suf[i] = suf[i + 1] + 1;
    } else {
      suf[i] = suf[i + 1];
    }
  }
  long long finalans = 0;
  for (long long i = 0; i < n; i++) {
    long long p = 0;
    if (i > 0) {
      p = pre[i - 1];
    }
    long long a1 = p;
    long long a2 = p;
    for (long long j = i; j < n; j++) {
      if (inp[j] == 1) {
        a1 = max(a1, a2) + 1;
      } else {
        a2++;
      }
      long long tempans = max(a1, a2);
      long long s = 0;
      if (j != n - 1) {
        s = suf[j + 1];
      }
      tempans += s;
      finalans = max(finalans, tempans);
    }
  }
  cout << finalans << '\n';
}
signed main() {
  ios_base::sync_with_stdio(false);
  cout << fixed << setprecision(20);
  cin.tie(NULL);
  long long t = 1;
  while (t--) {
    solve();
  }
  return 0;
}

```