#include <bits/stdc++.h>
using namespace std;
template <typename T>
inline T sqr(T a) {
  return a * a;
};
template <typename T>
void dprint(T begin, T end) {
  for (auto i = begin; i != end; i++) cerr << (*i) << " ";
  cerr << "\n";
};
const int maxn = 2e3 + 10;
const int d[] = {1, 2, 1, 2};
int n, a[maxn], dp[maxn][4];
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", a + i);
  for (int k = 0; k < 4; k++)
    for (int i = 0; i < n; i++) {
      if (i) dp[i][k] = dp[i - 1][k];
      if (d[k] == a[i]) {
        dp[i][k] = 1;
        for (int j = 0; j < i; j++)
          if (d[k] == a[j]) dp[i][k] = max(dp[i][k], dp[j][k] + 1);
        if (k) {
          for (int j = 0; j < i; j++)
            if (d[k - 1] == a[j]) dp[i][k] = max(dp[i][k], dp[j][k - 1] + 1);
        }
      }
    }
  int ans = 0;
  for (int k = 0; k < 4; k++) ans = max(dp[n - 1][k], ans);
  cout << ans << endl;
  return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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;
template <typename T>
inline T sqr(T a) {
  return a * a;
};
template <typename T>
void dprint(T begin, T end) {
  for (auto i = begin; i != end; i++) cerr << (*i) << " ";
  cerr << "\n";
};
const int maxn = 2e3 + 10;
const int d[] = {1, 2, 1, 2};
int n, a[maxn], dp[maxn][4];
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", a + i);
  for (int k = 0; k < 4; k++)
    for (int i = 0; i < n; i++) {
      if (i) dp[i][k] = dp[i - 1][k];
      if (d[k] == a[i]) {
        dp[i][k] = 1;
        for (int j = 0; j < i; j++)
          if (d[k] == a[j]) dp[i][k] = max(dp[i][k], dp[j][k] + 1);
        if (k) {
          for (int j = 0; j < i; j++)
            if (d[k - 1] == a[j]) dp[i][k] = max(dp[i][k], dp[j][k - 1] + 1);
        }
      }
    }
  int ans = 0;
  for (int k = 0; k < 4; k++) ans = max(dp[n - 1][k], ans);
  cout << ans << endl;
  return 0;
}

```