#include <bits/stdc++.h>
using namespace std;
long long a[100010], b[100010];
int main(int argc, char *argv[]) {
  int n;
  long long t, Sum = 0;
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%lld", &t);
    Sum += t;
  }
  for (int i = 1; i <= n; ++i) scanf("%lld", &b[i]);
  sort(b + 1, b + 1 + n);
  if (Sum <= (b[n] + b[n - 1]))
    puts("YES");
  else
    puts("NO");
  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;
long long a[100010], b[100010];
int main(int argc, char *argv[]) {
  int n;
  long long t, Sum = 0;
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%lld", &t);
    Sum += t;
  }
  for (int i = 1; i <= n; ++i) scanf("%lld", &b[i]);
  sort(b + 1, b + 1 + n);
  if (Sum <= (b[n] + b[n - 1]))
    puts("YES");
  else
    puts("NO");
  return 0;
}

```