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.
#include <bits/stdc++.h>
using namespace std;
const int N = 111;
struct node {
  int a, b;
  node() {}
  node(int a, int b) : a(a), b(b) {}
} a[N];
int cmp(node a, node b) {
  if (a.b != b.b) return a.b > b.b;
  return a.a > b.a;
}
int dp[2][N][N * N];
int main() {
  int n;
  while (cin >> n) {
    int sa = 0;
    for (int i = 1; i <= n; ++i) cin >> a[i].a, sa += a[i].a;
    for (int i = 1; i <= n; ++i) cin >> a[i].b;
    sort(a + 1, a + 1 + n, cmp);
    int k = 0, sb = 0;
    for (int i = 1; !k && i <= n; ++i) {
      sb += a[i].b;
      if (sb >= sa) k = i;
    }
    memset(dp, -1, sizeof(dp));
    int curb = 0;
    dp[0][0][0] = dp[1][0][0] = 0;
    for (int i = 1; i <= n; ++i) {
      curb += a[i].b;
      for (int j = 1; j <= k; ++j) {
        if (j > i) break;
        for (int w = curb; w >= 0; --w) {
          dp[i & 1][j][w] = dp[1 - (i & 1)][j][w];
          if (w < a[i].b || dp[1 - (i & 1)][j - 1][w - a[i].b] == -1) continue;
          dp[i & 1][j][w] =
              max(dp[1 - (i & 1)][j - 1][w - a[i].b] + a[i].a, dp[i & 1][j][w]);
        }
      }
    }
    int maxi = 0;
    for (int i = sa; i <= curb; ++i) maxi = max(maxi, dp[n & 1][k][i]);
    int t = sa - maxi;
    cout << k << " " << t << endl;
  }
  return 0;
}
