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;
long long BigMod(long long B, long long P, long long M) {
  long long R = 1;
  while (P > 0) {
    if (P % 2 == 1) {
      R = (R * B) % M;
    }
    P /= 2;
    B = (B * B) % M;
  }
  return R;
}
long long histogram(long long arr[], long long n) {
  stack<int> s;
  int area = 0, i = 0;
  while (i < n) {
    if (s.empty() || arr[s.top()] <= arr[i])
      s.push(i++);
    else {
      int tp = s.top();
      s.pop();
      int xx = arr[tp] * (s.empty() ? i : i - s.top() - 1);
      area = max(area, xx);
    }
  }
  while (!s.empty()) {
    int tp = s.top();
    s.pop();
    int xx = arr[tp] * (s.empty() ? i : i - s.top() - 1);
    area = max(area, xx);
  }
  return area;
}
map<long long, int> mp;
int main() {
  long long a, b;
  string s;
  long long arr[100000];
  cin >> a;
  long long sum = 0;
  for (int i = 0; i < a; i++) {
    cin >> b;
    sum += b;
  }
  long long p = 0;
  vector<int> v;
  for (int i = 0; i < a; i++) {
    cin >> b;
    v.push_back(b);
  }
  sort(v.rbegin(), v.rend());
  if ((v[0] + v[1]) >= sum)
    cout << "YES" << endl;
  else
    cout << "NO" << endl;
  return 0;
}
