#include <bits/stdc++.h>
using namespace std;
const int MAXN = 2 * 1e3 + 50;
int n;
int a[MAXN];
int pre[MAXN], suf[MAXN];
int f[MAXN][MAXN], g[MAXN][MAXN];
int ans;
int main() {
  scanf("%d", &n);
  for (register int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  for (register int i = 1; i <= n; ++i) pre[i] = pre[i - 1] + (a[i] == 1);
  for (register int i = n; i >= 1; --i) suf[i] = suf[i + 1] + (a[i] == 2);
  for (register int i = 1; i <= n; ++i)
    for (register int j = i; j <= n; ++j) {
      g[i][j] = g[i][j - 1] + (a[j] == 2);
      f[i][j] = max(f[i][j - 1], g[i][j - 1]) + (a[j] == 1);
    }
  for (register int i = 1; i <= n; ++i)
    for (register int j = i; j <= n; ++j)
      ans = max(ans, pre[i - 1] + suf[j + 1] + max(f[i][j], g[i][j]));
  printf("%d", ans);
  return 0;
}

### Prompt

Please provide a CPP coded solution 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;
const int MAXN = 2 * 1e3 + 50;
int n;
int a[MAXN];
int pre[MAXN], suf[MAXN];
int f[MAXN][MAXN], g[MAXN][MAXN];
int ans;
int main() {
  scanf("%d", &n);
  for (register int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  for (register int i = 1; i <= n; ++i) pre[i] = pre[i - 1] + (a[i] == 1);
  for (register int i = n; i >= 1; --i) suf[i] = suf[i + 1] + (a[i] == 2);
  for (register int i = 1; i <= n; ++i)
    for (register int j = i; j <= n; ++j) {
      g[i][j] = g[i][j - 1] + (a[j] == 2);
      f[i][j] = max(f[i][j - 1], g[i][j - 1]) + (a[j] == 1);
    }
  for (register int i = 1; i <= n; ++i)
    for (register int j = i; j <= n; ++j)
      ans = max(ans, pre[i - 1] + suf[j + 1] + max(f[i][j], g[i][j]));
  printf("%d", ans);
  return 0;
}

```