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;
int fn(long long q) {
  if (q <= 9) return q;
  int r = (q % 10);
  int w;
  int d = q;
  int s = 0;
  while (d > 0) {
    w = d;
    d /= 10;
  }
  if (w > r) {
    s -= 1;
  }
  s += (q / 10);
  s += 9;
  return s;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  long long n;
  cin >> n;
  long long a[n];
  long long b[n];
  long long s = 0;
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
    s += a[i];
  }
  for (long long i = 0; i < n; i++) cin >> b[i];
  sort(b, b + n);
  long long t = b[n - 1] + b[n - 2];
  if (t < s)
    cout << "NO";
  else
    cout << "YES";
  return 0;
}