#include <bits/stdc++.h>
using namespace std;
bool compare(int a, int b) { return a > b; }
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  cin >> t;
  int a[100009], b[100009];
  int i;
  for (i = 0; i < t; i++) cin >> a[i];
  for (i = 0; i < t; i++) cin >> b[i];
  long long sum = 0;
  sort(b, b + t, compare);
  for (i = 0; i < t; i++) sum += a[i];
  if ((b[0] + b[1]) < sum)
    cout << "NO" << endl;
  else
    cout << "YES" << endl;
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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".

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
bool compare(int a, int b) { return a > b; }
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  cin >> t;
  int a[100009], b[100009];
  int i;
  for (i = 0; i < t; i++) cin >> a[i];
  for (i = 0; i < t; i++) cin >> b[i];
  long long sum = 0;
  sort(b, b + t, compare);
  for (i = 0; i < t; i++) sum += a[i];
  if ((b[0] + b[1]) < sum)
    cout << "NO" << endl;
  else
    cout << "YES" << endl;
  return 0;
}

```