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 x, y;
} A[105];
int dp[10005][105];
bool cmp(node h, node g) { return h.y > g.y; }
bool cmm(node h, node g) { return h.x > g.x; }
int main() {
  int n, m, i, j, sum = 0, cnt = 0, ans = 0, q, ans2 = 0;
  cin >> n;
  for (i = 1; i <= n; i++) {
    scanf("%d", &A[i].x);
    sum += A[i].x;
  }
  for (i = 1; i <= n; i++) {
    scanf("%d", &A[i].y);
  }
  sort(A + 1, A + n + 1, cmp);
  for (i = 1; i < n; i++) {
    cnt += A[i].y;
    if (cnt >= sum) break;
  }
  int k = i;
  for (i = i + 1; i <= n; i++) {
    ans += A[i].x;
  }
  sort(A + 1, A + n + 1, cmm);
  for (i = 0; i <= 10000; i++) {
    for (j = 0; j <= 100; j++) {
      dp[i][j] = -1;
    }
  }
  dp[0][0] = 0;
  for (i = 1; i <= n; i++) {
    for (j = k - 1; j >= 0; j--) {
      for (q = cnt; q >= 0; q--) {
        if (q + A[i].y > cnt) continue;
        if (dp[q][j] != -1) {
          dp[q + A[i].y][j + 1] = max(dp[q + A[i].y][j + 1], dp[q][j] + A[i].x);
        }
      }
    }
  }
  for (i = sum; i <= cnt; i++) {
    if (sum - dp[i][k] < ans) {
      ans = sum - dp[i][k];
    }
  }
  printf("%d %d", k, ans);
  return 0;
}
