/*
Guess who has a bitmask dp solution
that will work on the first subtask?
Heh, that's right, STARLIGHT ANYA
Also, subtask 2 is cute because you just sort the gyms by X_i
and then find the longest prefix of X that works
All in all, 24 points ^^
I will O(n^2) later ^^
*/
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
vector<ll> x;
vector<ll> l;
ll n;
bool memo[1 << 10] = {false};
bool solved[1 << 10] = {false};
bool poss(int S) {
bool& ans = memo[S];
if(!solved[S]) {
if(S == 0) {
ans = true;
return ans;
}
ans = false;
ll sumx = 0;
for(int i = 0; i < 10; i++) {
sumx += ((S >> i) & 1 ? x[i] : 0);
}
for(int i = 0; i < 10; i++) {
if((S >> i) & 1) {
if(poss(S ^ (1 << i)) && (sumx - x[i]) <= l[i]) {
ans = true;
break;
}
}
}
}
solved[S] = true;
return ans;
}
ll subtask2_solve(vector<ll> x, ll l) {
sort(x.begin(), x.end());
ll sum = 0ll;
ll i = 0ll;
ll n = x.size();
while(i < n && sum <= l) {
sum += x[i];
i++;
}
return i;
}
int main() {
cin >> n;
for(ll i = 0ll; i < n; i++) {
ll cur;
cin >> cur;
x.push_back(cur);
}
for(ll i = 0ll; i < n; i++) {
ll cur;
cin >> cur;
l.push_back(cur);
}
if(n <= 10) {
poss((1 << n) - 1);
int ans = 0;
for(int i = 0; i < (1 << 10); i++) {
if(memo[i]) {
ans = max(ans, __popcount(i));
}
}
cout << ans;
} else {
// Assume subtask 2 constraints
cout << subtask2_solve(x, l[0]) << endl;
}
return 0;
}
