#include <bits/stdc++.h>
using namespace std;
int input[2002], n;
int ones[2002], twos[2002], ll[2002], rr[2002];
void prepare() {
  int i;
  ones[0] = input[0] == 1;
  twos[0] = 1 - ones[0];
  for (i = 1; i < n; i++) {
    ones[i] = ones[i - 1] + (input[i] == 1);
    twos[i] = i + 1 - ones[i];
  }
  for (i = 0; i < n; i++) {
    ll[i] = 0;
    rr[i] = 0;
  }
}
int getones(int l, int r) {
  if (l > 0)
    return ones[r] - ones[l - 1];
  else
    return ones[r];
}
int gettwos(int l, int r) {
  if (l > 0)
    return twos[r] - twos[l - 1];
  else
    return twos[r];
}
int main() {
  cin >> n;
  int i, j;
  for (__typeof(n) i = (0) - ((0) > (n)); i != (n) - ((0) > (n));
       i += 1 - 2 * ((0) > (n)))
    cin >> input[i];
  prepare();
  for (i = 0; i < n; i++)
    for (j = 0; j <= i; j++) ll[i] = max(ll[i], getones(0, j) + gettwos(j, i));
  for (i = 0; i < n; i++)
    for (j = i; j < n; j++)
      rr[i] = max(rr[i], getones(i, j) + gettwos(j, n - 1));
  int out = 0;
  for (i = 0; i < n - 1; i++) out = max(out, ll[i] + rr[i + 1]);
  out = max(out, ll[n - 1]);
  cout << out;
  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;
int input[2002], n;
int ones[2002], twos[2002], ll[2002], rr[2002];
void prepare() {
  int i;
  ones[0] = input[0] == 1;
  twos[0] = 1 - ones[0];
  for (i = 1; i < n; i++) {
    ones[i] = ones[i - 1] + (input[i] == 1);
    twos[i] = i + 1 - ones[i];
  }
  for (i = 0; i < n; i++) {
    ll[i] = 0;
    rr[i] = 0;
  }
}
int getones(int l, int r) {
  if (l > 0)
    return ones[r] - ones[l - 1];
  else
    return ones[r];
}
int gettwos(int l, int r) {
  if (l > 0)
    return twos[r] - twos[l - 1];
  else
    return twos[r];
}
int main() {
  cin >> n;
  int i, j;
  for (__typeof(n) i = (0) - ((0) > (n)); i != (n) - ((0) > (n));
       i += 1 - 2 * ((0) > (n)))
    cin >> input[i];
  prepare();
  for (i = 0; i < n; i++)
    for (j = 0; j <= i; j++) ll[i] = max(ll[i], getones(0, j) + gettwos(j, i));
  for (i = 0; i < n; i++)
    for (j = i; j < n; j++)
      rr[i] = max(rr[i], getones(i, j) + gettwos(j, n - 1));
  int out = 0;
  for (i = 0; i < n - 1; i++) out = max(out, ll[i] + rr[i + 1]);
  out = max(out, ll[n - 1]);
  cout << out;
  return 0;
}

```