#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:210886422")
using namespace std;
const int MOD = 998244353;
const int INF = 1000000001;
const long double EPS = 1e-13;
const int HASH_POW = 29;
const long double PI = acos(-1.0);
mt19937_64 rnd(chrono::high_resolution_clock::now().time_since_epoch().count());
long long nextLong(long long l, long long r) {
  uniform_int_distribution<long long> foo(l, r);
  return foo(rnd);
}
double nextDouble(double l, double r) {
  uniform_real_distribution<double> foo(l, r);
  return foo(rnd);
}
double workTime() { return double(clock()) / CLOCKS_PER_SEC; }
void myReturn(int code = 0) { exit(code); }
int n, a[2010], cnt1[2010], cnt2[2010];
int pref[2010], suff[2010];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
    if (a[i] == 1) {
      ++cnt1[i];
    } else {
      ++cnt2[i];
    }
  }
  for (int i = 2; i <= n + 1; ++i) {
    cnt1[i] += cnt1[i - 1];
    cnt2[i] += cnt2[i - 1];
  }
  for (int i = 1; i <= n; ++i) {
    for (int j = 0; j <= i; ++j) {
      pref[i] = max(pref[i], cnt1[j] + cnt2[i] - cnt2[j]);
    }
  }
  for (int i = n; i >= 1; --i) {
    for (int j = n; j >= i - 1; --j) {
      suff[i] = max(suff[i], cnt2[n] - cnt2[j] + cnt1[j] - cnt1[i - 1]);
    }
  }
  int ans = 0;
  for (int i = 0; i <= n; ++i) {
    ans = max(ans, pref[i] + suff[i + 1]);
  }
  cout << ans << endl;
  myReturn();
}

### 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>
#pragma comment(linker, "/STACK:210886422")
using namespace std;
const int MOD = 998244353;
const int INF = 1000000001;
const long double EPS = 1e-13;
const int HASH_POW = 29;
const long double PI = acos(-1.0);
mt19937_64 rnd(chrono::high_resolution_clock::now().time_since_epoch().count());
long long nextLong(long long l, long long r) {
  uniform_int_distribution<long long> foo(l, r);
  return foo(rnd);
}
double nextDouble(double l, double r) {
  uniform_real_distribution<double> foo(l, r);
  return foo(rnd);
}
double workTime() { return double(clock()) / CLOCKS_PER_SEC; }
void myReturn(int code = 0) { exit(code); }
int n, a[2010], cnt1[2010], cnt2[2010];
int pref[2010], suff[2010];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
    if (a[i] == 1) {
      ++cnt1[i];
    } else {
      ++cnt2[i];
    }
  }
  for (int i = 2; i <= n + 1; ++i) {
    cnt1[i] += cnt1[i - 1];
    cnt2[i] += cnt2[i - 1];
  }
  for (int i = 1; i <= n; ++i) {
    for (int j = 0; j <= i; ++j) {
      pref[i] = max(pref[i], cnt1[j] + cnt2[i] - cnt2[j]);
    }
  }
  for (int i = n; i >= 1; --i) {
    for (int j = n; j >= i - 1; --j) {
      suff[i] = max(suff[i], cnt2[n] - cnt2[j] + cnt1[j] - cnt1[i - 1]);
    }
  }
  int ans = 0;
  for (int i = 0; i <= n; ++i) {
    ans = max(ans, pref[i] + suff[i + 1]);
  }
  cout << ans << endl;
  myReturn();
}

```