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 namespace std;
void file() {
  freopen("in.txt", "r", stdin);
  freopen("out.txt", "w", stdout);
}
void Mohamed_Galal() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL), cout.tie(NULL);
}
int main() {
  long long n, sum1 = 0, sum2 = 0;
  cin >> n;
  int *arr = new int[n];
  int *arr2 = new int[n];
  for (size_t i = 0; i < n; i++) cin >> arr[i], sum1 += arr[i];
  ;
  for (size_t i = 0; i < n; i++) cin >> arr2[i];
  sort(arr2, arr2 + n);
  sum2 = arr2[n - 1] + arr2[n - 2];
  if (sum1 <= sum2 || n == 2)
    puts("YES");
  else
    puts("NO");
  return 0;
}
