#include <bits/stdc++.h>
using namespace std;
const double pi = acos(-1.0);
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const double EPS = 1e-9;
int n, a[2015];
int pre1[2015], suf2[2015], ans;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  for (int i = 1; i <= n; ++i) {
    pre1[i] = (1 == a[i] ? pre1[i - 1] + 1 : pre1[i - 1]);
  }
  for (int i = n; i; --i) {
    suf2[i] = (2 == a[i] ? suf2[i + 1] + 1 : suf2[i + 1]);
  }
  for (int i = 1; i <= n; ++i) {
    if (2 == a[i]) {
      ans = max(ans, pre1[i] + suf2[i]);
    }
  }
  for (int l = 1; l <= n; ++l) {
    int t1 = pre1[l - 1], ma = 0, cnt_2 = 0;
    for (int r = l; r <= n; ++r) {
      int t2 = suf2[r + 1];
      if (1 == a[r]) {
        ++ma;
      } else {
        ++cnt_2;
        ma = max(ma, cnt_2);
      }
      ans = max(ans, t1 + t2 + ma);
    }
  }
  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;
const double pi = acos(-1.0);
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const double EPS = 1e-9;
int n, a[2015];
int pre1[2015], suf2[2015], ans;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  for (int i = 1; i <= n; ++i) {
    pre1[i] = (1 == a[i] ? pre1[i - 1] + 1 : pre1[i - 1]);
  }
  for (int i = n; i; --i) {
    suf2[i] = (2 == a[i] ? suf2[i + 1] + 1 : suf2[i + 1]);
  }
  for (int i = 1; i <= n; ++i) {
    if (2 == a[i]) {
      ans = max(ans, pre1[i] + suf2[i]);
    }
  }
  for (int l = 1; l <= n; ++l) {
    int t1 = pre1[l - 1], ma = 0, cnt_2 = 0;
    for (int r = l; r <= n; ++r) {
      int t2 = suf2[r + 1];
      if (1 == a[r]) {
        ++ma;
      } else {
        ++cnt_2;
        ma = max(ma, cnt_2);
      }
      ans = max(ans, t1 + t2 + ma);
    }
  }
  printf("%d\n", ans);
  return 0;
}

```