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;
struct node {
  int ase, cap;
  node() {}
  node(int _koyta, int _pr) {
    ase = _koyta;
    cap = _pr;
  }
  bool operator<(const node& P) const {
    if (cap == P.cap) return ase < P.ase;
    return cap < P.cap;
  }
};
struct edge {
  int koyta, pr;
  edge() {}
  edge(int _koyta, int _pr) {
    koyta = _koyta;
    pr = _pr;
  }
  bool operator<(const edge& P) const {
    if (koyta == P.koyta) return pr < P.pr;
    return koyta < P.koyta;
  }
};
vector<node> vc;
int n;
edge dp[120][32000];
int vis[120][32000];
edge dp_func(int pos, int container) {
  if (pos == n) {
    if (container <= 20000) return {0, 0};
    return {200, 42000};
  }
  int l;
  if (vis[pos][container]) return dp[pos][container];
  vis[pos][container] = 1;
  edge ret = {200, 42000}, temp;
  temp = dp_func(pos + 1, container - vc[pos].cap);
  temp.koyta += 1;
  ret = min(ret, temp);
  temp = dp_func(pos + 1, container + vc[pos].ase);
  temp.pr += vc[pos].ase;
  ret = min(ret, temp);
  if (container > 20000) {
    l = container - 20000;
    l = min(l, vc[pos].cap);
    temp = dp_func(pos + 1, container - l);
    temp.koyta += 1;
    ret = min(ret, temp);
  }
  if (container < 20000) {
    l = 20000 - container;
    if (l < vc[pos].ase) {
      temp = dp_func(pos + 1, 20000 + vc[pos].ase - l);
      temp.pr += vc[pos].ase;
      ret = min(ret, temp);
    } else {
      temp = dp_func(pos + 1, container + vc[pos].ase);
      temp.pr += vc[pos].ase;
      ret = min(ret, temp);
    }
  }
  return dp[pos][container] = ret;
}
int main() {
  int i, j, k, l, m;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%d", &l);
    node gr;
    gr.ase = l;
    vc.push_back(gr);
  }
  for (i = 1; i <= n; i++) {
    scanf("%d", &l);
    vc[i - 1].cap = l - vc[i - 1].ase;
  }
  edge gr = dp_func(0, 20000);
  printf("%d %d\n", gr.koyta, gr.pr);
}
