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;
double const EPS = 1e-9, PI = acos(-1);
const int N = 1e2 + 9, M = 1e4 + 9, OO = 1e9 + 7, MOD = 1e9 + 7;
const long long inf = 1e18;
int a[N], b[N], mem[N][M], mem2[N][M], taken[N], total, n, sumBot = OO;
int solve(int idx, int sum) {
  if (idx == n) {
    if (sum >= total) return 0;
    return OO;
  }
  int& ret = mem[idx][sum];
  if (~ret) return ret;
  ret = OO;
  int c1 = solve(idx + 1, sum);
  int c2 = solve(idx + 1, sum + b[idx]) + 1;
  return ret = min(c1, c2);
}
int solve2(int idx, int sum) {
  if (idx == n) return 0;
  int& ret = mem2[idx][sum];
  if (~ret) return ret;
  ret = OO;
  int res = solve(idx, sum);
  int c1 = solve(idx + 1, sum);
  int c2 = solve(idx + 1, sum + b[idx]) + 1;
  if (c1 == c2 && c1 == res) {
    ret = min(solve2(idx + 1, sum) + a[idx], solve2(idx + 1, sum + b[idx]));
  } else if (c1 == res) {
    ret = solve2(idx + 1, sum) + a[idx];
  } else if (c2 == res) {
    ret = solve2(idx + 1, sum + b[idx]);
  } else
    assert(0);
  return ret;
}
int main() {
  cout << fixed << setprecision(12), ios::sync_with_stdio(false),
      cin.tie(nullptr), cout.tie(nullptr);
  cin >> n;
  for (int i = 0; i < n; ++i) cin >> a[i], total += a[i];
  for (int i = 0; i < n; ++i) cin >> b[i];
  memset(mem, -1, sizeof mem);
  memset(mem2, -1, sizeof mem2);
  cout << solve(0, 0) << ' ' << solve2(0, 0);
  return 0;
}
