#include <bits/stdc++.h>
using namespace std;
const long long inf = 1e18 + 7;
const long long N = 1e5 + 3;
const long long md = 1e9 + 7;
long long a[N], b[N];
long long n, s;
void sol() {
  s = 0;
  cin >> n;
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
    s += a[i];
  }
  for (long long i = 0; i < n; i++) {
    cin >> b[i];
  }
  long long mx1 = -1, mx2 = -1, mx1p, mx2p;
  for (long long i = 0; i < n; i++) {
    if (b[i] > mx1) {
      mx1 = b[i];
      mx1p = i;
    }
  }
  for (long long i = 0; i < n; i++) {
    if (i != mx1p) {
      if (b[i] > mx2) {
        mx2 = b[i];
        mx2p = i;
      }
    }
  }
  if (b[mx1p] + b[mx2p] >= s) {
    cout << "YES\n";
  } else
    cout << "NO\n";
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long T = 1;
  while (T--) sol();
  return 0;
}

### Prompt

Generate a cpp solution to the following 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".

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const long long inf = 1e18 + 7;
const long long N = 1e5 + 3;
const long long md = 1e9 + 7;
long long a[N], b[N];
long long n, s;
void sol() {
  s = 0;
  cin >> n;
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
    s += a[i];
  }
  for (long long i = 0; i < n; i++) {
    cin >> b[i];
  }
  long long mx1 = -1, mx2 = -1, mx1p, mx2p;
  for (long long i = 0; i < n; i++) {
    if (b[i] > mx1) {
      mx1 = b[i];
      mx1p = i;
    }
  }
  for (long long i = 0; i < n; i++) {
    if (i != mx1p) {
      if (b[i] > mx2) {
        mx2 = b[i];
        mx2p = i;
      }
    }
  }
  if (b[mx1p] + b[mx2p] >= s) {
    cout << "YES\n";
  } else
    cout << "NO\n";
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long T = 1;
  while (T--) sol();
  return 0;
}

```