#include <bits/stdc++.h>
using namespace std;
int n, a[102], b[102];
pair<int, int> dp[102][2 * 10002];
bool vis[102][2 * 10002];
pair<int, int> solve(int i, int diff) {
  if (i == n + 1)
    return diff >= 10000 ? pair<int, int>(0, 0) : pair<int, int>(1e9, 1e9);
  if (vis[i][diff]) return dp[i][diff];
  vis[i][diff] = true;
  pair<int, int> take = solve(i + 1, diff - a[i]);
  take.second += a[i];
  pair<int, int> leave = solve(i + 1, diff + b[i] - a[i]);
  ++leave.first;
  return dp[i][diff] = min(take, leave);
}
int main() {
  cin >> n;
  for (int i = (1); i <= (int)(n); ++i) scanf("%d", &a[i]);
  for (int i = (1); i <= (int)(n); ++i) scanf("%d", &b[i]);
  pair<int, int> ans = solve(1, 10000);
  cout << ans.first << " " << ans.second;
}

### Prompt

In CPP, your task is to solve the following problem:

Nick has n bottles of soda left after his birthday. Each bottle is described by two values: remaining amount of soda ai and bottle volume bi (ai ≤ bi).

Nick has decided to pour all remaining soda into minimal number of bottles, moreover he has to do it as soon as possible. Nick spends x seconds to pour x units of soda from one bottle to another.

Nick asks you to help him to determine k — the minimal number of bottles to store all remaining soda and t — the minimal time to pour soda into k bottles. A bottle can't store more soda than its volume. All remaining soda should be saved.

Input

The first line contains positive integer n (1 ≤ n ≤ 100) — the number of bottles.

The second line contains n positive integers a1, a2, ..., an (1 ≤ ai ≤ 100), where ai is the amount of soda remaining in the i-th bottle.

The third line contains n positive integers b1, b2, ..., bn (1 ≤ bi ≤ 100), where bi is the volume of the i-th bottle.

It is guaranteed that ai ≤ bi for any i.

Output

The only line should contain two integers k and t, where k is the minimal number of bottles that can store all the soda and t is the minimal time to pour the soda into k bottles.

Examples

Input

4
3 3 4 3
4 7 6 5


Output

2 6


Input

2
1 1
100 100


Output

1 1


Input

5
10 30 5 6 24
10 41 7 8 24


Output

3 11

Note

In the first example Nick can pour soda from the first bottle to the second bottle. It will take 3 seconds. After it the second bottle will contain 3 + 3 = 6 units of soda. Then he can pour soda from the fourth bottle to the second bottle and to the third bottle: one unit to the second and two units to the third. It will take 1 + 2 = 3 seconds. So, all the soda will be in two bottles and he will spend 3 + 3 = 6 seconds to do it.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int n, a[102], b[102];
pair<int, int> dp[102][2 * 10002];
bool vis[102][2 * 10002];
pair<int, int> solve(int i, int diff) {
  if (i == n + 1)
    return diff >= 10000 ? pair<int, int>(0, 0) : pair<int, int>(1e9, 1e9);
  if (vis[i][diff]) return dp[i][diff];
  vis[i][diff] = true;
  pair<int, int> take = solve(i + 1, diff - a[i]);
  take.second += a[i];
  pair<int, int> leave = solve(i + 1, diff + b[i] - a[i]);
  ++leave.first;
  return dp[i][diff] = min(take, leave);
}
int main() {
  cin >> n;
  for (int i = (1); i <= (int)(n); ++i) scanf("%d", &a[i]);
  for (int i = (1); i <= (int)(n); ++i) scanf("%d", &b[i]);
  pair<int, int> ans = solve(1, 10000);
  cout << ans.first << " " << ans.second;
}

```