#include <bits/stdc++.h>
int dx[] = {-1, -1, -1, 0, 1, 1, 1, 0};
int dy[] = {-1, 0, 1, 1, 1, 0, -1, -1};
using namespace std;
long long power(long long a, long long p, long long M) {
  long long r = 1;
  a = a % M;
  while (p > 0) {
    if (p % 2) r = (r * a) % M;
    a = (a * a) % M;
    p /= 2;
  }
  return r;
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long n;
long long arr[2005];
vector<pair<long long, long long> > one, zero;
int dp[2005][5][2005][2];
bool check[2005][5][2005][2];
long long solve(int ind, int c, long long co, bool last) {
  if (c == 4) return co - 1;
  if (ind >= n) return co;
  if (check[ind][c][co][last]) return dp[ind][c][co][last];
  check[ind][c][co][last] = 1;
  if (arr[ind] != last)
    return dp[ind][c][co][last] = max(solve(ind + 1, c + 1, co + 1, !last),
                                      solve(ind + 1, c, co, last));
  return dp[ind][c][co][last] =
             max(solve(ind + 1, c, co + 1, last), solve(ind + 1, c, co, last));
}
void solve() {
  cin >> n;
  for (long long i = 0; i < n; i++) {
    long long x;
    scanf("%lld", &x);
    arr[i] = x - 1;
  }
  cout << solve(0, 0, 0, 0) << endl;
}
int main() {
  solve();
  return 0;
}

### 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>
int dx[] = {-1, -1, -1, 0, 1, 1, 1, 0};
int dy[] = {-1, 0, 1, 1, 1, 0, -1, -1};
using namespace std;
long long power(long long a, long long p, long long M) {
  long long r = 1;
  a = a % M;
  while (p > 0) {
    if (p % 2) r = (r * a) % M;
    a = (a * a) % M;
    p /= 2;
  }
  return r;
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long n;
long long arr[2005];
vector<pair<long long, long long> > one, zero;
int dp[2005][5][2005][2];
bool check[2005][5][2005][2];
long long solve(int ind, int c, long long co, bool last) {
  if (c == 4) return co - 1;
  if (ind >= n) return co;
  if (check[ind][c][co][last]) return dp[ind][c][co][last];
  check[ind][c][co][last] = 1;
  if (arr[ind] != last)
    return dp[ind][c][co][last] = max(solve(ind + 1, c + 1, co + 1, !last),
                                      solve(ind + 1, c, co, last));
  return dp[ind][c][co][last] =
             max(solve(ind + 1, c, co + 1, last), solve(ind + 1, c, co, last));
}
void solve() {
  cin >> n;
  for (long long i = 0; i < n; i++) {
    long long x;
    scanf("%lld", &x);
    arr[i] = x - 1;
  }
  cout << solve(0, 0, 0, 0) << endl;
}
int main() {
  solve();
  return 0;
}

```