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 Tp>
inline void outarr(Tp _array, const std::size_t _N, const char* _delim = " ") {
  Tp end = _array + _N;
  for (Tp current = _array; current != end; ++current) {
    std::cout << *current << _delim;
  }
  std::cout << '\n';
}
using ll = std::int64_t;
using pii = std::pair<int, int>;
constexpr int INF = INT_MAX;
constexpr int MOD = 1e9 + 7;
int arr[100001];
int main() {
  std::ios_base::sync_with_stdio(false);
  std::cin.tie(nullptr);
  int n;
  cin >> n;
  ll sum = 0;
  for (int i = 0; i < (n); ++i) {
    int a;
    cin >> a;
    sum += a;
  }
  for (int i = 0; i < (n); ++i) {
    cin >> arr[i];
  }
  sort(arr, arr + n, [](const int a, const int b) { return b < a; });
  if ((ll)arr[0] + arr[1] >= sum) {
    return std::cout << ("YES"), 0;
    ;
  }
  cout << "NO";
  return 0;
}