#include <bits/stdc++.h>
using namespace std;
int dp[110][10010], n, s1, s2, ans, num, x;
struct node {
  int l, r;
} f[10010];
int cmp(node a, node b) {
  if (a.r != b.r)
    return a.r > b.r;
  else
    return a.l > b.l;
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &f[i].l);
    s1 += f[i].l;
  }
  for (int i = 1; i <= n; i++) {
    scanf("%d", &f[i].r);
    s2 += f[i].r;
  }
  x = s1;
  sort(f + 1, f + 1 + n, cmp);
  while ("Chtholly") {
    x -= f[++num].r;
    if (x <= 0) break;
  }
  memset(dp, -0x7f, sizeof dp);
  dp[0][0] = 0;
  for (int i = 1; i <= n; i++)
    for (int j = s2; j >= f[i].r; j--)
      for (int k = 1; k <= num; k++)
        dp[k][j] = max(dp[k][j], dp[k - 1][j - f[i].r] + f[i].l);
  for (int i = s1; i <= s2; i++) ans = max(ans, dp[num][i]);
  cout << num << " " << s1 - ans;
}

### Prompt

Please formulate 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 dp[110][10010], n, s1, s2, ans, num, x;
struct node {
  int l, r;
} f[10010];
int cmp(node a, node b) {
  if (a.r != b.r)
    return a.r > b.r;
  else
    return a.l > b.l;
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &f[i].l);
    s1 += f[i].l;
  }
  for (int i = 1; i <= n; i++) {
    scanf("%d", &f[i].r);
    s2 += f[i].r;
  }
  x = s1;
  sort(f + 1, f + 1 + n, cmp);
  while ("Chtholly") {
    x -= f[++num].r;
    if (x <= 0) break;
  }
  memset(dp, -0x7f, sizeof dp);
  dp[0][0] = 0;
  for (int i = 1; i <= n; i++)
    for (int j = s2; j >= f[i].r; j--)
      for (int k = 1; k <= num; k++)
        dp[k][j] = max(dp[k][j], dp[k - 1][j - f[i].r] + f[i].l);
  for (int i = s1; i <= s2; i++) ans = max(ans, dp[num][i]);
  cout << num << " " << s1 - ans;
}

```