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 long N = 110;
long long n, m, sum, cnt, p, f[N][N * N], ans;
struct data {
  long long x, y;
} a[N];
bool cmp(data x, data y) { return x.y == y.y ? x.x > y.x : x.y > y.y; }
signed main() {
  scanf("%lld", &n);
  for (long long i = 1; i <= n; i++) scanf("%lld", &a[i].x), sum += a[i].x;
  for (long long i = 1; i <= n; i++) scanf("%lld", &a[i].y), m += a[i].y;
  sort(a + 1, a + 1 + n, cmp);
  for (long long i = 1; i <= n; i++) {
    cnt += a[i].y;
    if (cnt >= sum) {
      p = i;
      break;
    }
  }
  memset(f, -0x3f, sizeof(f)), f[0][0] = 0;
  for (long long i = 1; i <= n; i++)
    for (long long j = m; j >= a[i].y; j--)
      for (long long k = 1; k <= p; k++)
        f[k][j] = max(f[k][j], f[k - 1][j - a[i].y] + a[i].x);
  for (long long i = sum; i <= m; i++) ans = max(ans, f[p][i]);
  printf("%lld %lld\n", p, sum - ans);
  return 0;
}
