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 INF = 0x3f3f3f3f;
const long long INFF = 1e18;
const double pi = acos(-1.0);
const double inf = 1e18;
const double eps = 1e-9;
const long long mod = 1e9 + 7;
const int maxmat = 10;
const unsigned long long BASE = 133333331;
inline void RI(int &x) {
  char c;
  while ((c = getchar()) < '0' || c > '9')
    ;
  x = c - '0';
  while ((c = getchar()) >= '0' && c <= '9') x = (x << 3) + (x << 1) + c - '0';
}
typedef struct node {
  int left, volume;
} node;
node a[105];
bool cmp(node a, node b) {
  if (a.volume != b.volume)
    return a.volume > b.volume;
  else
    return a.left > b.left;
}
int main() {
  int n;
  while (~scanf("%d", &n)) {
    int dp[105][10005];
    int sum = 0;
    memset(dp, -1, sizeof(dp));
    for (int i = 1; i <= n; i++) {
      scanf("%d", &a[i].left);
      sum += a[i].left;
    }
    for (int i = 1; i <= n; i++) scanf("%d", &a[i].volume);
    int cnt = 0;
    int total = 0;
    sort(a + 1, a + 1 + n, cmp);
    for (int i = 1; i <= n; i++) {
      total += a[i].volume;
      if (total >= sum) {
        cnt = i;
        break;
      }
    }
    dp[0][0] = 0;
    for (int i = 1; i <= n; i++) {
      for (int j = sum; (j - a[i].left) >= 0; j--) {
        for (int k = i; k >= 1; k--) {
          if (dp[k - 1][j - a[i].left] != -1)
            dp[k][j] = max(dp[k][j], dp[k - 1][j - a[i].left] + a[i].volume);
        }
      }
    }
    int ans = 0;
    for (int i = sum; i >= 1; i--) {
      if (dp[cnt][i] >= sum) {
        ans = sum - i;
        break;
      }
    }
    printf("%d %d\n", cnt, ans);
  }
}
