#include <bits/stdc++.h>
#pragma GCC Optimize("Ofast")
using namespace std;
const long long maxn = 2e3 + 7;
const long long mod = 1e9 + 7;
const long long INF = 3e18 + 315;
long long n, a[maxn], dp[maxn][5];
template <typename... Args>
void input(Args&... args) {
  ((cin >> args), ...);
}
template <typename... Args>
void output(Args&... args) {
  ((cout << args), ...);
}
void no();
void yes();
int32_t main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  input(n);
  for (long long i = 1; i <= n; i++) input(a[i]);
  for (long long i = 1; i <= n; i++)
    for (long long j = 0; j < 4; j++) {
      ((dp[i][j]) = max(dp[i][j], dp[i - 1][j]));
      if (j > 0) ((dp[i][j]) = max(dp[i][j], dp[i - 1][j - 1]));
      (!(j & 1) ? dp[i][j] += a[i] == 1 : dp[i][j] += a[i] == 2);
    }
  long long ans = -INF;
  for (long long i = 0; i < 4; i++) ((ans) = max(ans, dp[n][i]));
  output(ans, "\n");
  return 0;
}
void no() { cout << "NO" << endl; }
void yes() { cout << "YES" << endl; }

### Prompt

Create a solution in cpp for 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>
#pragma GCC Optimize("Ofast")
using namespace std;
const long long maxn = 2e3 + 7;
const long long mod = 1e9 + 7;
const long long INF = 3e18 + 315;
long long n, a[maxn], dp[maxn][5];
template <typename... Args>
void input(Args&... args) {
  ((cin >> args), ...);
}
template <typename... Args>
void output(Args&... args) {
  ((cout << args), ...);
}
void no();
void yes();
int32_t main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  input(n);
  for (long long i = 1; i <= n; i++) input(a[i]);
  for (long long i = 1; i <= n; i++)
    for (long long j = 0; j < 4; j++) {
      ((dp[i][j]) = max(dp[i][j], dp[i - 1][j]));
      if (j > 0) ((dp[i][j]) = max(dp[i][j], dp[i - 1][j - 1]));
      (!(j & 1) ? dp[i][j] += a[i] == 1 : dp[i][j] += a[i] == 2);
    }
  long long ans = -INF;
  for (long long i = 0; i < 4; i++) ((ans) = max(ans, dp[n][i]));
  output(ans, "\n");
  return 0;
}
void no() { cout << "NO" << endl; }
void yes() { cout << "YES" << endl; }

```