Problem: Jafar has n cans of cola. Each can is described by two integers: remaining volume of cola ai and can's capacity bi (ai ≤  bi).

Jafar has decided to pour all remaining cola into just 2 cans, determine if he can do this or not!

Input

The first line of the input contains one integer n (2 ≤ n ≤ 100 000) — number of cola cans.

The second line contains n space-separated integers a1, a2, ..., an (0 ≤ ai ≤ 109) — volume of remaining cola in cans.

The third line contains n space-separated integers that b1, b2, ..., bn (ai ≤ bi ≤ 109) — capacities of the cans.

Output

Print "YES" (without quotes) if it is possible to pour all remaining cola in 2 cans. Otherwise print "NO" (without quotes).

You can print each letter in any case (upper or lower).

Examples

Input

2
3 5
3 6


Output

YES


Input

3
6 8 9
6 10 12


Output

NO


Input

5
0 0 5 0 0
1 1 8 10 5


Output

YES


Input

4
4 1 0 3
5 2 2 3


Output

YES

Note

In the first sample, there are already 2 cans, so the answer is "YES".

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <typename P, typename Q>
ostream& operator<<(ostream& os, pair<P, Q> p) {
  os << "(" << p.first << "," << p.second << ")";
  return os;
}
int main(int argc, char* argv[]) {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  while (cin >> n) {
    vector<long long int> a(n);
    vector<long long int> b(n);
    for (auto& i : a) cin >> i;
    for (auto& i : b) cin >> i;
    long long int sum = accumulate(a.begin(), a.end(), 0LL);
    sort(b.begin(), b.end());
    reverse(b.begin(), b.end());
    cout << (b[0] + b[1] < sum ? "NO" : "YES") << endl;
  }
  return 0;
}