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;
int gcd(int x, int y) {
  if (y == 0) return x;
  return gcd(y, x % y);
}
int lcm(int x, int y) { return x / gcd(x, y) * y; }
int lowbit(int x) { return x & (-x); }
long long quickmod(long long x, long long y, long long mod) {
  x = x % mod;
  long long ans = 1;
  while (y) {
    if (y & 1) ans = ans * x % mod;
    y >>= 1;
    x = x * x % mod;
  }
  return ans;
}
const int INF = 0x3f3f3f3f;
const double EPS = 1e-7;
const double PI = acos(-1.0);
const long long MOD = 1e9 + 7;
int n;
int num1[105], num2[105];
int dp[2][105][20005];
int main() {
  while (scanf("%d", &n) != EOF) {
    memset(dp, -1, sizeof(dp));
    int sum = 0;
    for (int i = 1; i <= n; i++) {
      scanf("%d", &num1[i]);
      sum += num1[i];
    }
    for (int i = 1; i <= n; i++) scanf("%d", &num2[i]);
    dp[0][0][0] = 0;
    int flag = 0;
    for (int i = 1; i <= n; i++) {
      flag ^= 1;
      for (int j = 0; j <= i; j++) {
        for (int k = 0; k <= 20000; k++) {
          dp[flag][j][k] = dp[flag ^ 1][j][k];
          if (j >= 1 && k >= num2[i] && dp[flag ^ 1][j - 1][k - num2[i]] != -1)
            dp[flag][j][k] =
                max(dp[flag][j][k], dp[flag ^ 1][j - 1][k - num2[i]] + num1[i]);
        }
      }
    }
    int ans1 = -1, ans2 = 0;
    for (int j = 1; j <= n; j++) {
      ans2 = 0;
      for (int k = 20000; k >= sum; k--) {
        if (dp[flag][j][k] != -1) {
          ans1 = j;
          ans2 = max(ans2, dp[flag][j][k]);
        }
      }
      if (ans1 != -1) break;
    }
    printf("%d %d\n", ans1, sum - ans2);
  }
  return 0;
}
