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, i, j, flag = 0;
  long long s = 0;
  cin >> n;
  int a[n + 5], b[n + 5];
  for (i = 0; i < n; i++) {
    cin >> a[i];
  }
  for (i = 0; i < n; i++) {
    cin >> b[i];
  }
  vector<pair<int, int>> v1, v;
  for (i = 0; i < n; i++) {
    v.push_back(make_pair(a[i], b[i] - a[i]));
  }
  sort(v.begin(), v.end());
  for (i = 0; i < n - 2; i++) {
    s = s + v[i].first;
  }
  int y = v[n - 1].second + v[n - 2].second;
  if (y >= s)
    cout << "YES";
  else
    flag = 1;
  if (flag == 1) {
    for (i = 0; i < n; i++) {
      v1.push_back(make_pair(b[i] - a[i], a[i]));
    }
    sort(v1.begin(), v1.end());
    long long p = 0;
    for (i = 0; i < n - 2; i++) {
      p = p + v1[i].second;
    }
    int m = v1[n - 2].first + v1[n - 1].first;
    if (m >= p)
      cout << "YES";
    else
      cout << "NO";
  }
  cout << endl;
}
