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;
template <typename T>
bool bitcheck(T p, T pos) {
  return (bool)(p & (1 << pos));
}
template <typename T>
T biton(T p, T pos) {
  return p = p | (1 << pos);
}
template <typename T>
T bitoff(T p, T pos) {
  return p = p & ~(1 << pos);
}
template <typename T>
long long int toint(T s) {
  long long int p;
  stringstream ss(s);
  ss >> p;
  return p;
}
template <typename T>
string tostring(T n) {
  stringstream ss;
  ss << n;
  return ss.str();
}
template <typename T>
T POW(T b, T p) {
  T Ans = 1;
  while (p) {
    if (p & 1) Ans = (Ans * b);
    b = (b * b);
    p >>= 1;
  }
  return Ans;
}
template <typename T>
T BigMod(T b, T p, T Mod) {
  T Ans = 1;
  while (p) {
    if (p & 1) Ans = (Ans * b) % Mod;
    b = (b * b) % Mod;
    p >>= 1;
  }
  return Ans;
}
template <typename T>
T ModInverse(T p, T Mod) {
  return BigMod(p, Mod - 2, Mod);
}
const int fx[] = {+0, +0, +1, -1, -1, +1, -1, +1};
const int fy[] = {-1, +1, +0, +0, +1, +1, -1, -1};
const int mx = 1e5 + 10;
const long long int inf = (long long int)1e9;
const int mod = 1e9 + 7;
const double pi = 2 * acos(0.0);
const double E = 2.71828182845904523536;
const double eps = 1e-9;
bool mark[mx];
vector<int> prime;
void sieve() {
  memset(mark, true, sizeof(mark));
  for (int i = 3; i <= (int)sqrt(mx); i += 2) {
    if (mark[i]) {
      for (int j = i * i; j <= mx; j += 2 * i) mark[j] = false;
    }
  }
  prime.push_back(2);
  for (int i = 3; i <= mx; i += 2)
    if (mark[i]) prime.push_back(i);
}
long long int a[110], total, n;
vector<pair<long long int, long long int> > v;
pair<long long int, long long int> dp[110][10010];
bool vis[110][10010];
pair<long long int, long long int> boss(long long int i, long long int s) {
  if (i == n && s < total) return {inf, inf};
  if (i == n && s >= total) return {0, 0};
  if (vis[i][s]) return dp[i][s];
  vis[i][s] = 1;
  pair<long long int, long long int> p1 =
      make_pair(1 + boss(i + 1, s + v[i].first).first,
                boss(i + 1, s + v[i].first).second);
  pair<long long int, long long int> p2 =
      make_pair(boss(i + 1, s).first, v[i].second + boss(i + 1, s).second);
  return dp[i][s] = min(p1, p2);
}
int main() {
  total = 0;
  cin >> n;
  for (int i = 0; i <= n - 1; i++) {
    cin >> a[i];
    total += a[i];
  }
  for (int i = 0; i <= n - 1; i++) {
    long long int k;
    cin >> k;
    v.push_back({k, a[i]});
  }
  pair<long long int, long long int> d = boss(0, 0);
  cout << d.first << " " << d.second << '\n';
  return 0;
}
