#include <bits/stdc++.h>
using namespace std;
const int N = 105;
const int MX = N * N;
const int INF = ~0x3f3f3f3f;
int a[N], b[N];
int dp[N][MX];
int main() {
  int n;
  while (scanf("%d", &n) == 1) {
    int sum = 0;
    for (int i = 0; i < n; i++) {
      scanf("%d", a + i);
      sum += a[i];
    }
    for (int i = 0; i < n; i++) {
      scanf("%d", b + i);
    }
    memset(dp, INF, sizeof dp);
    dp[0][0] = 0;
    for (int i = 0; i < n; i++) {
      int aa = a[i], bb = b[i];
      for (int j = i; j >= 0; j--) {
        for (int k = 0; k < MX; k++) {
          if (dp[j][k] != INF) {
            dp[j + 1][k + bb] = max(dp[j + 1][k + bb], dp[j][k] + aa);
          }
        }
      }
    }
    for (int i = 1; i <= n; i++) {
      int mx = INF;
      for (int j = sum; j < MX; j++) {
        mx = max(mx, dp[i][j]);
      }
      if (mx != INF) {
        printf("%d %d\n", i, sum - mx);
        break;
      }
    }
  }
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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 N = 105;
const int MX = N * N;
const int INF = ~0x3f3f3f3f;
int a[N], b[N];
int dp[N][MX];
int main() {
  int n;
  while (scanf("%d", &n) == 1) {
    int sum = 0;
    for (int i = 0; i < n; i++) {
      scanf("%d", a + i);
      sum += a[i];
    }
    for (int i = 0; i < n; i++) {
      scanf("%d", b + i);
    }
    memset(dp, INF, sizeof dp);
    dp[0][0] = 0;
    for (int i = 0; i < n; i++) {
      int aa = a[i], bb = b[i];
      for (int j = i; j >= 0; j--) {
        for (int k = 0; k < MX; k++) {
          if (dp[j][k] != INF) {
            dp[j + 1][k + bb] = max(dp[j + 1][k + bb], dp[j][k] + aa);
          }
        }
      }
    }
    for (int i = 1; i <= n; i++) {
      int mx = INF;
      for (int j = sum; j < MX; j++) {
        mx = max(mx, dp[i][j]);
      }
      if (mx != INF) {
        printf("%d %d\n", i, sum - mx);
        break;
      }
    }
  }
  return 0;
}

```