#include <bits/stdc++.h>
using namespace std;
const int N = 2222;
struct node {
  int val, cnt1, cnt2;
  node(int e, int x, int y) {
    val = e;
    cnt1 = x;
    cnt2 = y;
  }
  node() { val = cnt1 = cnt2 = 0; }
};
node operator+(const node &x, const node &y) {
  node sum;
  sum.cnt1 = x.cnt1 + y.cnt1;
  sum.cnt2 = x.cnt2 + y.cnt2;
  sum.val = max(x.val + y.cnt1, x.cnt2 + y.val);
  return sum;
}
int n, a[N];
int f1[N], f2[N];
node T[4 * N];
void build(int p, int l, int r) {
  if (l > r) return;
  if (l == r) {
    node tmp;
    if (a[l] == 1) {
      tmp.val = 1;
      tmp.cnt1 = 1;
      tmp.cnt2 = 0;
    } else {
      tmp.val = 1;
      tmp.cnt1 = 0;
      tmp.cnt2 = 1;
    }
    T[p] = tmp;
    return;
  }
  int mid = (l + r) / 2;
  build(2 * p, l, mid);
  build(2 * p + 1, mid + 1, r);
  T[p] = T[2 * p] + T[2 * p + 1];
}
node get(int p, int l, int r, int i, int j) {
  if (i > r || j < l) {
    node tmp;
    return tmp;
  }
  if (l >= i && r <= j) {
    return T[p];
  }
  int mid = (l + r) / 2;
  return get(2 * p, l, mid, i, j) + get(2 * p + 1, mid + 1, r, i, j);
}
int main(int argc, char const *argv[]) {
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
  }
  for (int i = 1; i <= n; ++i) {
    f1[i] = f1[i - 1];
    if (a[i] == 1) {
      f1[i] = f1[i - 1] + 1;
    }
  }
  for (int i = n; i; --i) {
    f2[i] = f2[i + 1];
    if (a[i] == 2) {
      f2[i] = f2[i + 1] + 1;
    }
  }
  build(1, 1, n);
  int get_max = 0;
  for (int i = 1; i <= n; ++i) {
    get_max = max(get_max, f1[i] + f2[i]);
  }
  for (int i = 1; i <= n; ++i) {
    for (int j = i; j <= n; ++j) {
      int res = f1[i - 1] + f2[j + 1] + get(1, 1, n, i, j).val;
      get_max = max(get_max, res);
    }
  }
  cout << get_max;
  return 0;
}

### Prompt

In Cpp, your task is to solve 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 int N = 2222;
struct node {
  int val, cnt1, cnt2;
  node(int e, int x, int y) {
    val = e;
    cnt1 = x;
    cnt2 = y;
  }
  node() { val = cnt1 = cnt2 = 0; }
};
node operator+(const node &x, const node &y) {
  node sum;
  sum.cnt1 = x.cnt1 + y.cnt1;
  sum.cnt2 = x.cnt2 + y.cnt2;
  sum.val = max(x.val + y.cnt1, x.cnt2 + y.val);
  return sum;
}
int n, a[N];
int f1[N], f2[N];
node T[4 * N];
void build(int p, int l, int r) {
  if (l > r) return;
  if (l == r) {
    node tmp;
    if (a[l] == 1) {
      tmp.val = 1;
      tmp.cnt1 = 1;
      tmp.cnt2 = 0;
    } else {
      tmp.val = 1;
      tmp.cnt1 = 0;
      tmp.cnt2 = 1;
    }
    T[p] = tmp;
    return;
  }
  int mid = (l + r) / 2;
  build(2 * p, l, mid);
  build(2 * p + 1, mid + 1, r);
  T[p] = T[2 * p] + T[2 * p + 1];
}
node get(int p, int l, int r, int i, int j) {
  if (i > r || j < l) {
    node tmp;
    return tmp;
  }
  if (l >= i && r <= j) {
    return T[p];
  }
  int mid = (l + r) / 2;
  return get(2 * p, l, mid, i, j) + get(2 * p + 1, mid + 1, r, i, j);
}
int main(int argc, char const *argv[]) {
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
  }
  for (int i = 1; i <= n; ++i) {
    f1[i] = f1[i - 1];
    if (a[i] == 1) {
      f1[i] = f1[i - 1] + 1;
    }
  }
  for (int i = n; i; --i) {
    f2[i] = f2[i + 1];
    if (a[i] == 2) {
      f2[i] = f2[i + 1] + 1;
    }
  }
  build(1, 1, n);
  int get_max = 0;
  for (int i = 1; i <= n; ++i) {
    get_max = max(get_max, f1[i] + f2[i]);
  }
  for (int i = 1; i <= n; ++i) {
    for (int j = i; j <= n; ++j) {
      int res = f1[i - 1] + f2[j + 1] + get(1, 1, n, i, j).val;
      get_max = max(get_max, res);
    }
  }
  cout << get_max;
  return 0;
}

```