#include <bits/stdc++.h>
using namespace std;
int read() {
  int x;
  scanf("%d", &x);
  return x;
}
int n, lat;
int a[2005];
int last[2005];
int sum2[2005];
int sum1[2005];
int f[2005][2005];
int g[2005][2005];
int ans;
void readin() {
  n = read();
  for (int i = 1; i <= n; i++) {
    a[i] = read(), last[i] = lat, f[i][i] = 1, g[i][i] = 1;
    if (a[i] == 2)
      lat = i, sum2[i] = sum2[i - 1] + 1, sum1[i] = sum1[i - 1];
    else
      sum2[i] = sum2[i - 1], sum1[i] = sum1[i - 1] + 1;
  }
  for (int i = 0; i <= n; i++) {
    ans = max(ans, sum1[i] + sum2[n] - sum2[i]);
  }
  for (int i = 1; i <= n; i++) {
    for (int j = i + 1; j <= n; j++) {
      if (a[j] == 1)
        f[i][j] = g[i][j - 1] + 1, g[i][j] = f[i][j];
      else
        f[i][j] = f[i][last[j]] + 1, g[i][j] = max(g[i][j - 1], f[i][j]);
      ans = max(ans, sum1[i - 1] + sum2[n] - sum2[j] + g[i][j]);
    }
  }
  printf("%d\n", ans);
}
int main() {
  readin();
  return 0;
}

### Prompt

Generate 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;
int read() {
  int x;
  scanf("%d", &x);
  return x;
}
int n, lat;
int a[2005];
int last[2005];
int sum2[2005];
int sum1[2005];
int f[2005][2005];
int g[2005][2005];
int ans;
void readin() {
  n = read();
  for (int i = 1; i <= n; i++) {
    a[i] = read(), last[i] = lat, f[i][i] = 1, g[i][i] = 1;
    if (a[i] == 2)
      lat = i, sum2[i] = sum2[i - 1] + 1, sum1[i] = sum1[i - 1];
    else
      sum2[i] = sum2[i - 1], sum1[i] = sum1[i - 1] + 1;
  }
  for (int i = 0; i <= n; i++) {
    ans = max(ans, sum1[i] + sum2[n] - sum2[i]);
  }
  for (int i = 1; i <= n; i++) {
    for (int j = i + 1; j <= n; j++) {
      if (a[j] == 1)
        f[i][j] = g[i][j - 1] + 1, g[i][j] = f[i][j];
      else
        f[i][j] = f[i][last[j]] + 1, g[i][j] = max(g[i][j - 1], f[i][j]);
      ans = max(ans, sum1[i - 1] + sum2[n] - sum2[j] + g[i][j]);
    }
  }
  printf("%d\n", ans);
}
int main() {
  readin();
  return 0;
}

```