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 = 110;
const int INF = 0x3f3f3f3f;
struct node {
  int v, r;
  bool operator<(const node& u) const { return v > u.v; }
} a[N];
int dp[110][10010];
int sum[N];
int main() {
  int n, tot = 0, k = 0;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i].r), tot += a[i].r;
  for (int i = 1; i <= n; i++) scanf("%d", &a[i].v);
  sort(a + 1, a + n + 1);
  for (int i = 1; i <= n; i++) sum[i] = sum[i - 1] + a[i].v;
  for (int i = 1; i <= n; i++)
    if (sum[i] >= tot) {
      k = i;
      break;
    }
  memset(dp, INF, sizeof(dp));
  dp[0][0] = 0;
  for (int i = 1; i <= n; i++)
    for (int j = min(i - 1, k); j >= 0; j--)
      for (int x = sum[i - 1]; x >= 0; x--) {
        dp[j + 1][x + a[i].v] = min(dp[j][x], dp[j + 1][x + a[i].v]);
        if (dp[j][x] != INF) dp[j][x] += a[i].r;
      }
  int ans = INF;
  for (int i = tot; i <= sum[k]; i++) ans = min(ans, dp[k][i]);
  printf("%d %d\n", k, ans);
  return 0;
}
