#include <bits/stdc++.h>
using namespace std;
inline int read() {
  char c = getchar();
  int x = 0, f = 1;
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') x = (x << 3) + (x << 1) + c - '0', c = getchar();
  return x * f;
}
int n, sum1[2010], sum2[2010], a[2010], ans;
int main() {
  n = read();
  for (int i = 1; i <= n; ++i) a[i] = read();
  for (int i = 1; i <= n; ++i) sum1[i] = sum1[i - 1] + (a[i] == 1);
  for (int i = n; i >= 1; --i) sum2[i] = sum2[i + 1] + (a[i] == 2);
  for (int pos = 1; pos <= n + 1; ++pos) {
    int ans1 = 0, ans2 = 0;
    for (int l = 1; l <= pos; ++l) {
      ans1 = max(sum1[l - 1] + sum2[l] - sum2[pos], ans1);
    }
    for (int r = pos; r <= n + 1; ++r) {
      ans2 = max(sum2[r] + sum1[r - 1] - sum1[pos - 1], ans2);
    }
    ans = max(ans, ans1 + ans2);
  }
  printf("%d\n", ans);
  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;
inline int read() {
  char c = getchar();
  int x = 0, f = 1;
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') x = (x << 3) + (x << 1) + c - '0', c = getchar();
  return x * f;
}
int n, sum1[2010], sum2[2010], a[2010], ans;
int main() {
  n = read();
  for (int i = 1; i <= n; ++i) a[i] = read();
  for (int i = 1; i <= n; ++i) sum1[i] = sum1[i - 1] + (a[i] == 1);
  for (int i = n; i >= 1; --i) sum2[i] = sum2[i + 1] + (a[i] == 2);
  for (int pos = 1; pos <= n + 1; ++pos) {
    int ans1 = 0, ans2 = 0;
    for (int l = 1; l <= pos; ++l) {
      ans1 = max(sum1[l - 1] + sum2[l] - sum2[pos], ans1);
    }
    for (int r = pos; r <= n + 1; ++r) {
      ans2 = max(sum2[r] + sum1[r - 1] - sum1[pos - 1], ans2);
    }
    ans = max(ans, ans1 + ans2);
  }
  printf("%d\n", ans);
  return 0;
}

```