#include <bits/stdc++.h>
using namespace std;
const int N = 2000 + 5;
int n, v[N];
int a1[N];
int a2[N];
int dp21[N][N];
int dp[N];
int ap(int x, int st, int dr) {
  if (st > dr) return 0;
  if (x == 1)
    return a1[dr] - a1[st - 1];
  else
    return a2[dr] - a2[st - 1];
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> v[i];
    if (v[i] == 1)
      a1[i] = 1;
    else
      a2[i] = 1;
    a1[i] += a1[i - 1];
    a2[i] += a2[i - 1];
  }
  for (int st = 1; st <= n; st++) {
    for (int dr = st; dr <= n; dr++) {
      if (v[dr] == 2)
        dp21[st][dr] = dp21[st][dr - 1];
      else {
        dp21[st][dr] = max(ap(2, st, dr), dp21[st][dr - 1]) + 1;
      }
    }
  }
  int ans = 0;
  for (int st = 1; st <= n; st++) {
    for (int dr = st; dr <= n; dr++) {
      ans = max(ans, ap(1, 1, st - 1) + dp21[st][dr] + ap(2, dr + 1, n));
    }
  }
  for (int i = 1; i <= n; i++) {
    if (v[i] == 1)
      dp[i] = dp[i - 1];
    else {
      dp[i] = max(dp[i - 1], ap(1, 1, i)) + 1;
    }
  }
  ans = max(ans, dp[n]);
  cout << ans << "\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 int N = 2000 + 5;
int n, v[N];
int a1[N];
int a2[N];
int dp21[N][N];
int dp[N];
int ap(int x, int st, int dr) {
  if (st > dr) return 0;
  if (x == 1)
    return a1[dr] - a1[st - 1];
  else
    return a2[dr] - a2[st - 1];
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> v[i];
    if (v[i] == 1)
      a1[i] = 1;
    else
      a2[i] = 1;
    a1[i] += a1[i - 1];
    a2[i] += a2[i - 1];
  }
  for (int st = 1; st <= n; st++) {
    for (int dr = st; dr <= n; dr++) {
      if (v[dr] == 2)
        dp21[st][dr] = dp21[st][dr - 1];
      else {
        dp21[st][dr] = max(ap(2, st, dr), dp21[st][dr - 1]) + 1;
      }
    }
  }
  int ans = 0;
  for (int st = 1; st <= n; st++) {
    for (int dr = st; dr <= n; dr++) {
      ans = max(ans, ap(1, 1, st - 1) + dp21[st][dr] + ap(2, dr + 1, n));
    }
  }
  for (int i = 1; i <= n; i++) {
    if (v[i] == 1)
      dp[i] = dp[i - 1];
    else {
      dp[i] = max(dp[i - 1], ap(1, 1, i)) + 1;
    }
  }
  ans = max(ans, dp[n]);
  cout << ans << "\n";
  return 0;
}

```