#include <bits/stdc++.h>
using namespace std;
inline void read(int &x) {
  int f = 0;
  x = 0;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = 1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
  if (f) x = -x;
}
int a[2005], n, s1[2005], s2[2005], ans;
int main() {
  int i, j, k;
  read(n);
  for (i = 1; i <= n; i++) {
    read(a[i]);
    s1[i] = s1[i - 1] + (a[i] == 1);
    s2[i] = s2[i - 1] + (a[i] == 2);
  }
  for (i = 1; i <= n; i++) {
    int sa = 0, sb = 0;
    for (j = i; j >= 0; j--) sa = max(sa, s1[j] + s2[i] - s2[j]);
    for (j = i; j <= n; j++) sb = max(sb, s1[j] - s1[i] + s2[n] - s2[j]);
    ans = max(ans, sa + sb);
  }
  printf("%d", ans);
}

### Prompt

Develop a solution in Cpp 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;
inline void read(int &x) {
  int f = 0;
  x = 0;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = 1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
  if (f) x = -x;
}
int a[2005], n, s1[2005], s2[2005], ans;
int main() {
  int i, j, k;
  read(n);
  for (i = 1; i <= n; i++) {
    read(a[i]);
    s1[i] = s1[i - 1] + (a[i] == 1);
    s2[i] = s2[i - 1] + (a[i] == 2);
  }
  for (i = 1; i <= n; i++) {
    int sa = 0, sb = 0;
    for (j = i; j >= 0; j--) sa = max(sa, s1[j] + s2[i] - s2[j]);
    for (j = i; j <= n; j++) sb = max(sb, s1[j] - s1[i] + s2[n] - s2[j]);
    ans = max(ans, sa + sb);
  }
  printf("%d", ans);
}

```