#include <bits/stdc++.h>
using namespace std;
const int MAXN = 2005;
int n, a[MAXN], f1[MAXN], d[MAXN][MAXN];
int s2[MAXN << 2 | 1], s1[MAXN << 2], s[MAXN << 2];
struct Node {
  int s1, s2, s;
  Node(int s1 = 0, int s2 = 0, int s = 0) : s1(s1), s2(s2), s(s) {}
};
void build(int id, int l, int r) {
  if (l == r) {
    if (a[l] == 1)
      s1[id] = 1;
    else
      s2[id] = 1;
    s[id] = 1;
    return;
  }
  int mid = (l + r) >> 1;
  build(id << 1, l, mid);
  build(id << 1 | 1, mid + 1, r);
  s2[id] = s2[id << 1] + s2[id << 1 | 1];
  s1[id] = s1[id << 1] + s1[id << 1 | 1];
  s[id] = max(s[id << 1] + s1[id << 1 | 1], s2[id << 1] + s[id << 1 | 1]);
}
Node query(int id, int l, int r, int ql, int qr) {
  if (l == ql && r == qr) return Node(s1[id], s2[id], s[id]);
  int mid = (l + r) >> 1;
  if (qr <= mid) return query(id << 1, l, mid, ql, qr);
  if (ql > mid) return query(id << 1 | 1, mid + 1, r, ql, qr);
  Node lt = query(id << 1, l, mid, ql, mid),
       rt = query(id << 1 | 1, mid + 1, r, mid + 1, qr);
  return Node(lt.s1 + rt.s1, lt.s2 + rt.s2, max(lt.s + rt.s1, lt.s2 + rt.s));
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    if (a[i] == 1)
      f1[i] = f1[i - 1] + 1;
    else
      f1[i] = f1[i - 1];
  }
  build(1, 1, n);
  int ans = 0;
  for (int l = 1; l <= n; l++)
    for (int r = l; r <= n; r++) {
      Node p = query(1, 1, n, l, r);
      ans = max(ans, f1[l - 1] + p.s + n - r - f1[n] + f1[r]);
    }
  cout << ans << endl;
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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 = 2005;
int n, a[MAXN], f1[MAXN], d[MAXN][MAXN];
int s2[MAXN << 2 | 1], s1[MAXN << 2], s[MAXN << 2];
struct Node {
  int s1, s2, s;
  Node(int s1 = 0, int s2 = 0, int s = 0) : s1(s1), s2(s2), s(s) {}
};
void build(int id, int l, int r) {
  if (l == r) {
    if (a[l] == 1)
      s1[id] = 1;
    else
      s2[id] = 1;
    s[id] = 1;
    return;
  }
  int mid = (l + r) >> 1;
  build(id << 1, l, mid);
  build(id << 1 | 1, mid + 1, r);
  s2[id] = s2[id << 1] + s2[id << 1 | 1];
  s1[id] = s1[id << 1] + s1[id << 1 | 1];
  s[id] = max(s[id << 1] + s1[id << 1 | 1], s2[id << 1] + s[id << 1 | 1]);
}
Node query(int id, int l, int r, int ql, int qr) {
  if (l == ql && r == qr) return Node(s1[id], s2[id], s[id]);
  int mid = (l + r) >> 1;
  if (qr <= mid) return query(id << 1, l, mid, ql, qr);
  if (ql > mid) return query(id << 1 | 1, mid + 1, r, ql, qr);
  Node lt = query(id << 1, l, mid, ql, mid),
       rt = query(id << 1 | 1, mid + 1, r, mid + 1, qr);
  return Node(lt.s1 + rt.s1, lt.s2 + rt.s2, max(lt.s + rt.s1, lt.s2 + rt.s));
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    if (a[i] == 1)
      f1[i] = f1[i - 1] + 1;
    else
      f1[i] = f1[i - 1];
  }
  build(1, 1, n);
  int ans = 0;
  for (int l = 1; l <= n; l++)
    for (int r = l; r <= n; r++) {
      Node p = query(1, 1, n, l, r);
      ans = max(ans, f1[l - 1] + p.s + n - r - f1[n] + f1[r]);
    }
  cout << ans << endl;
  return 0;
}

```