#include <bits/stdc++.h>
using namespace std;
int readint() {
  int x;
  scanf("%d", &x);
  return x;
}
int dp[10005][105];
int main() {
  int n = readint();
  vector<int> a(n + 1), b(n + 1);
  vector<int> c(n + 1);
  int sum = 0;
  for (int i = 1; i <= n; i++) a[i] = readint(), sum += a[i];
  for (int i = 1; i <= n; i++) b[i] = readint(), c[i] = b[i];
  sort(c.begin(), c.end());
  reverse(c.begin() + 1, c.end());
  for (int i = 1; i <= n; i++) c[i] = c[i - 1] + c[i];
  int k = lower_bound(c.begin(), c.end(), sum) - c.begin();
  cout << k << ' ';
  memset(dp, 128, sizeof dp);
  dp[0][0] = 0;
  for (int i = 1; i <= n; i++)
    for (int j = c[k]; j >= b[i]; j--)
      for (int kk = 1; kk <= k; kk++)
        dp[j][kk] = max(dp[j][kk], dp[j - b[i]][kk - 1] + a[i]);
  int ans = -1;
  for (int i = sum; i <= c[k]; i++) ans = max(ans, dp[i][k]);
  cout << sum - ans;
}

### Prompt

Your challenge is to write a cpp solution to 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 readint() {
  int x;
  scanf("%d", &x);
  return x;
}
int dp[10005][105];
int main() {
  int n = readint();
  vector<int> a(n + 1), b(n + 1);
  vector<int> c(n + 1);
  int sum = 0;
  for (int i = 1; i <= n; i++) a[i] = readint(), sum += a[i];
  for (int i = 1; i <= n; i++) b[i] = readint(), c[i] = b[i];
  sort(c.begin(), c.end());
  reverse(c.begin() + 1, c.end());
  for (int i = 1; i <= n; i++) c[i] = c[i - 1] + c[i];
  int k = lower_bound(c.begin(), c.end(), sum) - c.begin();
  cout << k << ' ';
  memset(dp, 128, sizeof dp);
  dp[0][0] = 0;
  for (int i = 1; i <= n; i++)
    for (int j = c[k]; j >= b[i]; j--)
      for (int kk = 1; kk <= k; kk++)
        dp[j][kk] = max(dp[j][kk], dp[j - b[i]][kk - 1] + a[i]);
  int ans = -1;
  for (int i = sum; i <= c[k]; i++) ans = max(ans, dp[i][k]);
  cout << sum - ans;
}

```