#include <bits/stdc++.h>
using namespace std;
const double PI = acos(-1.0);
double fRand(double fMin, double fMax) {
  double f = (double)rand() / RAND_MAX;
  return fMin + f * (fMax - fMin);
}
template <class T>
T min(T a, T b, T c) {
  return min(a, min(b, c));
}
template <class T>
T max(T a, T b, T c) {
  return max(a, max(b, c));
}
int n, a[2005], d[3][2005], fl[2005], fr[2005];
int main() {
  scanf("%d", &n);
  for (int i = (1); i <= (n); ++i) {
    scanf("%d", &a[i]);
    for (int j = (1); j <= (2); ++j) d[j][i] = d[j][i - 1];
    ++d[a[i]][i];
  }
  for (int i = (1); i <= (n); ++i) {
    for (int j = (0); j <= (i); ++j)
      fl[i] = max(fl[i], d[1][j] + d[2][i] - d[2][j]);
  }
  for (int i = (1); i <= (n); ++i) {
    for (int j = (i); j <= (n); ++j)
      fr[i] = max(fr[i], d[2][n] - d[2][j] + d[1][j] - d[1][i - 1]);
  }
  int ans = 0;
  for (int i = (0); i <= (n); ++i) ans = max(ans, fl[i] + fr[i + 1]);
  printf("%d", ans);
  return 0;
}

### Prompt

In cpp, your task is to solve 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(-1.0);
double fRand(double fMin, double fMax) {
  double f = (double)rand() / RAND_MAX;
  return fMin + f * (fMax - fMin);
}
template <class T>
T min(T a, T b, T c) {
  return min(a, min(b, c));
}
template <class T>
T max(T a, T b, T c) {
  return max(a, max(b, c));
}
int n, a[2005], d[3][2005], fl[2005], fr[2005];
int main() {
  scanf("%d", &n);
  for (int i = (1); i <= (n); ++i) {
    scanf("%d", &a[i]);
    for (int j = (1); j <= (2); ++j) d[j][i] = d[j][i - 1];
    ++d[a[i]][i];
  }
  for (int i = (1); i <= (n); ++i) {
    for (int j = (0); j <= (i); ++j)
      fl[i] = max(fl[i], d[1][j] + d[2][i] - d[2][j]);
  }
  for (int i = (1); i <= (n); ++i) {
    for (int j = (i); j <= (n); ++j)
      fr[i] = max(fr[i], d[2][n] - d[2][j] + d[1][j] - d[1][i - 1]);
  }
  int ans = 0;
  for (int i = (0); i <= (n); ++i) ans = max(ans, fl[i] + fr[i + 1]);
  printf("%d", ans);
  return 0;
}

```