#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 50;
const double eps = 1e-8;
const int inf = 0x3f3f3f3f;
const long long lnf = 0x3f3f3f3f3f3f3f3f;
const long long mod = 998244353;
struct node {
  int a, b;
} p[105];
int dp[105][10050];
bool cmp(node A, node B) { return A.b > B.b; }
int main() {
  int n;
  int sum = 0;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &p[i].a);
    sum += p[i].a;
  }
  for (int i = 1; i <= n; i++) scanf("%d", &p[i].b);
  int cnt = 0;
  sort(p + 1, p + 1 + n, cmp);
  int temp = sum;
  for (int i = 1; i <= n; i++) {
    if (temp > 0) {
      temp -= p[i].b;
      cnt++;
    } else
      break;
  }
  memset(dp, -inf, sizeof(dp));
  dp[0][0] = 0;
  printf("%d ", cnt);
  for (int i = 1; i <= n; i++) {
    for (int j = sum; j >= p[i].a; j--) {
      for (int k = i; k >= 1; k--) {
        if (dp[k - 1][j - p[i].a] != -inf)
          dp[k][j] = max(dp[k][j], dp[k - 1][j - p[i].a] + p[i].b);
      }
    }
  }
  for (int i = sum; i >= 1; i--) {
    if (dp[cnt][i] >= sum) {
      printf("%d\n", sum - i);
      return 0;
    }
  }
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 50;
const double eps = 1e-8;
const int inf = 0x3f3f3f3f;
const long long lnf = 0x3f3f3f3f3f3f3f3f;
const long long mod = 998244353;
struct node {
  int a, b;
} p[105];
int dp[105][10050];
bool cmp(node A, node B) { return A.b > B.b; }
int main() {
  int n;
  int sum = 0;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &p[i].a);
    sum += p[i].a;
  }
  for (int i = 1; i <= n; i++) scanf("%d", &p[i].b);
  int cnt = 0;
  sort(p + 1, p + 1 + n, cmp);
  int temp = sum;
  for (int i = 1; i <= n; i++) {
    if (temp > 0) {
      temp -= p[i].b;
      cnt++;
    } else
      break;
  }
  memset(dp, -inf, sizeof(dp));
  dp[0][0] = 0;
  printf("%d ", cnt);
  for (int i = 1; i <= n; i++) {
    for (int j = sum; j >= p[i].a; j--) {
      for (int k = i; k >= 1; k--) {
        if (dp[k - 1][j - p[i].a] != -inf)
          dp[k][j] = max(dp[k][j], dp[k - 1][j - p[i].a] + p[i].b);
      }
    }
  }
  for (int i = sum; i >= 1; i--) {
    if (dp[cnt][i] >= sum) {
      printf("%d\n", sum - i);
      return 0;
    }
  }
  return 0;
}

```