#include <bits/stdc++.h>
using namespace std;
const double PI = acos(-1);
const int mod = 1e9 + 7;
const int MAX = 1000005;
int n;
vector<int> v;
int dp[4][2005];
int go(int ind, int sw) {
  if (ind == n) return 0;
  if (dp[sw][ind] != -1) return dp[sw][ind];
  int ret = go(ind + 1, sw);
  if (sw == 0) {
    if (v[ind] == 1)
      ret++;
    else
      ret = max(ret, go(ind + 1, 1) + 1);
  } else if (sw == 1) {
    if (v[ind] == 1)
      ret = max(ret, go(ind + 1, 2) + 1);
    else
      ret++;
  } else if (sw == 2) {
    if (v[ind] == 1)
      ret++;
    else
      ret = max(ret, go(ind + 1, 3) + 1);
  } else if (sw == 3) {
    if (v[ind] == 2) ret++;
  }
  return dp[sw][ind] = ret;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n;
  v.resize(n);
  for (int i = 0; i < n; i++) cin >> v[i];
  memset(dp, -1, sizeof dp);
  cout << go(0, 0) << "\n";
  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>
using namespace std;
const double PI = acos(-1);
const int mod = 1e9 + 7;
const int MAX = 1000005;
int n;
vector<int> v;
int dp[4][2005];
int go(int ind, int sw) {
  if (ind == n) return 0;
  if (dp[sw][ind] != -1) return dp[sw][ind];
  int ret = go(ind + 1, sw);
  if (sw == 0) {
    if (v[ind] == 1)
      ret++;
    else
      ret = max(ret, go(ind + 1, 1) + 1);
  } else if (sw == 1) {
    if (v[ind] == 1)
      ret = max(ret, go(ind + 1, 2) + 1);
    else
      ret++;
  } else if (sw == 2) {
    if (v[ind] == 1)
      ret++;
    else
      ret = max(ret, go(ind + 1, 3) + 1);
  } else if (sw == 3) {
    if (v[ind] == 2) ret++;
  }
  return dp[sw][ind] = ret;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n;
  v.resize(n);
  for (int i = 0; i < n; i++) cin >> v[i];
  memset(dp, -1, sizeof dp);
  cout << go(0, 0) << "\n";
  return 0;
}

```