#include <bits/stdc++.h>
using namespace std;
void __print(int x) { cerr << x; }
void __print(long x) { cerr << x; }
void __print(long long x) { cerr << x; }
void __print(unsigned x) { cerr << x; }
void __print(unsigned long x) { cerr << x; }
void __print(unsigned long long x) { cerr << x; }
void __print(float x) { cerr << x; }
void __print(double x) { cerr << x; }
void __print(long double x) { cerr << x; }
void __print(char x) { cerr << '\'' << x << '\''; }
void __print(const char *x) { cerr << '\"' << x << '\"'; }
void __print(const string &x) { cerr << '\"' << x << '\"'; }
void __print(bool x) { cerr << (x ? "true" : "false"); }
template <typename T, typename V>
void __print(const pair<T, V> &x) {
  cerr << '{';
  __print(x.first);
  cerr << ',';
  __print(x.second);
  cerr << '}';
}
template <typename T>
void __print(const T &x) {
  int f = 0;
  cerr << '{';
  for (auto &i : x) cerr << (f++ ? "," : ""), __print(i);
  cerr << "}";
}
void _print() { cerr << "]\n"; }
template <typename T, typename... V>
void _print(T t, V... v) {
  __print(t);
  if (sizeof...(v)) cerr << ", ";
  _print(v...);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int t = 1;
  while (t--) {
    long long int n = 0, m = 0, k = 0, i = 0, j = 0, p = 0, q = 0, x = 0, y = 0,
                  z = 0, ans = 0, cnt = 0, l = 0, r = 0, mid = 0, lo = 0,
                  hi = 0;
    string s;
    bool flag = false;
    cin >> n;
    long long int a[n];
    for (i = 0; i <= n - 1; i++) cin >> a[i];
    long long int dp[4];
    memset(dp, 0, sizeof(dp));
    for (i = 0; i <= n - 1; i++) {
      dp[3] = max(dp[3], dp[2]) + (a[i] == 2);
      dp[2] = max(dp[2], dp[1]) + (a[i] == 1);
      dp[1] = max(dp[1], dp[0]) + (a[i] == 2);
      dp[0] = dp[0] + (a[i] == 1);
    }
    ans = max(dp[0], dp[1]);
    ans = max(dp[2], ans);
    ans = max(dp[3], ans);
    cout << ans << endl;
    ;
  }
  return 0;
}

### Prompt

Your task is to create 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;
void __print(int x) { cerr << x; }
void __print(long x) { cerr << x; }
void __print(long long x) { cerr << x; }
void __print(unsigned x) { cerr << x; }
void __print(unsigned long x) { cerr << x; }
void __print(unsigned long long x) { cerr << x; }
void __print(float x) { cerr << x; }
void __print(double x) { cerr << x; }
void __print(long double x) { cerr << x; }
void __print(char x) { cerr << '\'' << x << '\''; }
void __print(const char *x) { cerr << '\"' << x << '\"'; }
void __print(const string &x) { cerr << '\"' << x << '\"'; }
void __print(bool x) { cerr << (x ? "true" : "false"); }
template <typename T, typename V>
void __print(const pair<T, V> &x) {
  cerr << '{';
  __print(x.first);
  cerr << ',';
  __print(x.second);
  cerr << '}';
}
template <typename T>
void __print(const T &x) {
  int f = 0;
  cerr << '{';
  for (auto &i : x) cerr << (f++ ? "," : ""), __print(i);
  cerr << "}";
}
void _print() { cerr << "]\n"; }
template <typename T, typename... V>
void _print(T t, V... v) {
  __print(t);
  if (sizeof...(v)) cerr << ", ";
  _print(v...);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int t = 1;
  while (t--) {
    long long int n = 0, m = 0, k = 0, i = 0, j = 0, p = 0, q = 0, x = 0, y = 0,
                  z = 0, ans = 0, cnt = 0, l = 0, r = 0, mid = 0, lo = 0,
                  hi = 0;
    string s;
    bool flag = false;
    cin >> n;
    long long int a[n];
    for (i = 0; i <= n - 1; i++) cin >> a[i];
    long long int dp[4];
    memset(dp, 0, sizeof(dp));
    for (i = 0; i <= n - 1; i++) {
      dp[3] = max(dp[3], dp[2]) + (a[i] == 2);
      dp[2] = max(dp[2], dp[1]) + (a[i] == 1);
      dp[1] = max(dp[1], dp[0]) + (a[i] == 2);
      dp[0] = dp[0] + (a[i] == 1);
    }
    ans = max(dp[0], dp[1]);
    ans = max(dp[2], ans);
    ans = max(dp[3], ans);
    cout << ans << endl;
    ;
  }
  return 0;
}

```