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 long double PI = 3.141592653589793238462643383279;
const long double EPS = 1e-7;
const long long INF = 1e9 + 7;
const long long MOD = INF;
const long long maxn = 2e5 + 1;
int a[101];
int b[101];
pair<int, int> dp[101][20001];
bool used[101][20001];
pair<int, int> calc(int i, int capac) {
  if (!i) {
    if (capac <= 10000) return make_pair(0, 0);
    return make_pair(1e9, 0);
  }
  if (used[i][capac]) return dp[i][capac];
  used[i][capac] = true;
  pair<int, int> p1, p2;
  p1 = calc(i - 1, capac - (b[i] - a[i]));
  p1.first += 1;
  p2 = calc(i - 1, capac + a[i]);
  p2.second += a[i];
  return dp[i][capac] = min(p1, p2);
}
int main() {
  int n;
  cin >> n;
  for (long long i = 1; i < n + 1; ++i) cin >> a[i];
  for (long long i = 1; i < n + 1; ++i) cin >> b[i];
  pair<int, int> res = calc(n, 10000);
  cout << res.first << " " << res.second;
}
