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;
inline int nextint() {
  int temporary;
  scanf("%d", &temporary);
  return temporary;
}
inline long long nextll() {
  long long temporary;
  scanf("%lld", &temporary);
  return temporary;
}
inline double nextdouble() {
  double temporary;
  scanf("%lf", &temporary);
  return temporary;
}
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long lcm(long long a, long long b) { return a * (b / gcd(a, b)); }
int leftChild(int p) { return p << 1; }
int rightChild(int p) { return (p << 1) + 1; }
inline long long mid(long long l, long long r) { return (l + r) / 2; }
const long long MOD = 1000000007;
const double eps = 1e-7;
const double PI = acos(-1.0);
int main() {
  std::ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n;
  cin >> n;
  long long rem = 0;
  long long a[n], b[n];
  for (long long i = (long long)(0); i < (long long)(n); i++) {
    cin >> a[i];
    rem += a[i];
  }
  for (long long i = (long long)(0); i < (long long)(n); i++) {
    cin >> b[i];
  }
  sort(b, b + n);
  if (b[n - 1] + b[n - 2] >= rem) {
    cout << "YES";
  } else {
    cout << "NO";
  }
  return 0;
}