#include <bits/stdc++.h>
using namespace std;
#pragma warning(disable : 4996)
#pragma comment(linker, "/STACK:336777216")
const int sqrtn = 300;
const int inf = 987654321;
const int sz = 1e5 + 5;
const int mod = 1e9 + 7;
const int N = 2005;
int n, a[N], s[N];
int one(int st, int en) {
  if (st > en) return 0;
  if (st == 0) return 0;
  return s[en] - s[st - 1];
}
int two(int st, int en) { return (en - st + 1) - one(st, en); }
int calc(int le, int ri) {
  int rlt = 0;
  for (int i = le - 1; i <= ri; i++) {
    rlt = max(rlt, one(le, i) + two(i + 1, ri));
  }
  return rlt;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i <= n; i++) {
    s[i] = s[i - 1] + (a[i] == 1);
  }
  int ans = 0;
  ans = calc(1, n);
  for (int i = 1; i < n; i++) {
    ans = max(ans, calc(1, i) + calc(i + 1, n));
  }
  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;
#pragma warning(disable : 4996)
#pragma comment(linker, "/STACK:336777216")
const int sqrtn = 300;
const int inf = 987654321;
const int sz = 1e5 + 5;
const int mod = 1e9 + 7;
const int N = 2005;
int n, a[N], s[N];
int one(int st, int en) {
  if (st > en) return 0;
  if (st == 0) return 0;
  return s[en] - s[st - 1];
}
int two(int st, int en) { return (en - st + 1) - one(st, en); }
int calc(int le, int ri) {
  int rlt = 0;
  for (int i = le - 1; i <= ri; i++) {
    rlt = max(rlt, one(le, i) + two(i + 1, ri));
  }
  return rlt;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i <= n; i++) {
    s[i] = s[i - 1] + (a[i] == 1);
  }
  int ans = 0;
  ans = calc(1, n);
  for (int i = 1; i < n; i++) {
    ans = max(ans, calc(1, i) + calc(i + 1, n));
  }
  printf("%d\n", ans);
  return 0;
}

```