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;
int main() {
  int n;
  cin >> n;
  long long int x = 0;
  long long int a, b;
  long long int maxi;
  long long int c[100000], d[100000];
  for (int j = 0; j < n; j++) {
    cin >> a;
    c[j] = a;
    x += a;
  }
  long long int l;
  for (int i = 0; i < n; i++) {
    cin >> b;
    d[i] = b;
    if (i == 0) {
      maxi = b;
      l = i;
    } else {
      if (maxi < b) {
        maxi = b;
        l = i;
      }
    }
  }
  long long int maxii;
  for (int i = 0; i < n; i++) {
    if (i == l) {
      continue;
    }
    if (i == 0) {
      maxii = d[i];
    } else {
      if (maxii < d[i]) {
        maxii = d[i];
      }
    }
  }
  int v = maxi + maxii;
  if (v >= x) {
    cout << "YES";
  } else {
    cout << "NO";
  }
  return 0;
}
