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;
bool cmp(int a, int b) { return a > b; }
int main() {
  int n, i;
  scanf("%d", &n);
  long long a[100005], sum = 0, b[100005];
  int max1 = 0, max2 = 0;
  for (i = 0; i < n; i++) {
    scanf("%lld", &a[i]);
    sum += a[i];
  }
  for (i = 0; i < n; i++) {
    scanf("%lld", &b[i]);
    if (b[i] > max1) {
      max2 = max1;
      max1 = b[i];
    } else if (b[i] > max2) {
      max2 = b[i];
    }
  }
  if (max1 + max2 >= sum)
    printf("YES");
  else
    printf("NO");
  return 0;
}