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".
#include <bits/stdc++.h>
using pair = std::pair<long, long>;
int main() {
  long n;
  std::cin >> n;
  n *= 2;
  std::vector<long> cans;
  while (n--) {
    long c;
    std::cin >> c;
    cans.push_back(c);
  }
  long sz = cans.size() / 2;
  std::vector<pair> cans_cap;
  for (int x = 0; x < sz; ++x) {
    cans_cap.push_back({cans[x], cans[x + sz]});
  }
  std::sort(cans_cap.begin(), cans_cap.end(),
            [](pair a, pair b) { return a.second > b.second; });
  long long vol = 0;
  for (auto p : cans_cap) {
    vol += p.first;
  }
  long a = cans_cap[0].second;
  long b = cans_cap[1].second;
  if ((a + b) >= vol) {
    std::cout << "YES\n";
  } else {
    std::cout << "NO\n";
  }
}
