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>
const int maxn = 110;
using namespace std;
struct bott {
  int wt;
  int bt;
} pp[maxn];
int cc[maxn];
bool cmp(bott a, bott b) { return a.bt > b.bt; }
int dp[maxn][10005];
int ans, n, wts, bts, mini, totb;
int main() {
  scanf("%d", &n);
  wts = 0, bts = 0;
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &pp[i].wt);
    wts += pp[i].wt;
  }
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &pp[i].bt);
    bts += pp[i].bt;
    cc[i] = pp[i].bt;
  }
  sort(cc + 1, cc + 1 + n);
  int temp = 0;
  for (int i = 1; i <= n && temp < wts; ++i) {
    temp += cc[n - i + 1];
    mini++;
  }
  memset(dp, -1, sizeof(dp));
  dp[0][0] = 0, ans = wts;
  for (int i = 1; i <= n; ++i)
    for (int j = bts; j >= pp[i].bt; --j)
      for (int l = mini; l; l--)
        if (dp[l - 1][j - pp[i].bt] >= 0)
          dp[l][j] = max(dp[l][j], dp[l - 1][j - pp[i].bt] + pp[i].wt);
  for (int j = bts; j >= wts; --j) ans = min(ans, wts - dp[mini][j]);
  printf("%d %d\n", mini, ans);
  return 0;
}
