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 MOD = 1e9 + 7;
const int INF = 2e9;
const int magic = 348;
const double eps = 1e-10;
struct bottle {
  int cur, vol;
} a[108];
int n, k, sum;
int dp[2][101][10048];
bool cmp(bottle first, bottle second) { return first.vol > second.vol; }
int main() {
  int i, j, m;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%d", &a[i].cur);
    sum += a[i].cur;
  }
  for (i = 1; i <= n; i++) scanf("%d", &a[i].vol);
  sort(a + 1, a + n + 1, cmp);
  int cc = 0;
  for (i = 1; i <= n; i++) {
    cc += a[i].vol;
    if (cc >= sum) break;
  }
  k = i;
  for (i = 0; i <= 1; i++)
    for (j = 0; j <= 100; j++)
      for (m = 0; m <= 10000; m++) dp[i][j][m] = -INF;
  dp[0][0][0] = 0;
  int ans = 0;
  int p0 = 0, p1 = 1;
  for (i = 1; i <= n; i++) {
    for (j = 0; j <= k; j++)
      for (m = 0; m <= 100 * i; m++) {
        dp[p1][j][m] = dp[p0][j][m];
        if (j && m - a[i].vol >= 0)
          dp[p1][j][m] =
              max(dp[p1][j][m], dp[p0][j - 1][m - a[i].vol] + a[i].cur);
        if (i == n && j == k && m >= sum) ans = max(ans, dp[p1][j][m]);
      }
    p0 ^= 1;
    p1 ^= 1;
  }
  printf("%d %d\n", k, sum - ans);
  return 0;
}
