#include <bits/stdc++.h>
using namespace std;
void Merge(int *a, int low, int high, int mid) {
  int i, j, k, temp[high - low + 1];
  i = low;
  k = 0;
  j = mid + 1;
  while (i <= mid && j <= high) {
    if (a[i] < a[j]) {
      temp[k] = a[i];
      k++;
      i++;
    } else {
      temp[k] = a[j];
      k++;
      j++;
    }
  }
  while (i <= mid) {
    temp[k] = a[i];
    k++;
    i++;
  }
  while (j <= high) {
    temp[k] = a[j];
    k++;
    j++;
  }
  for (i = low; i <= high; i++) {
    a[i] = temp[i - low];
  }
}
void MergeSort(int *a, int low, int high) {
  int mid;
  if (low < high) {
    mid = (low + high) / 2;
    MergeSort(a, low, mid);
    MergeSort(a, mid + 1, high);
    Merge(a, low, high, mid);
  }
}
int main() {
  long int i, n, sum = 0;
  cin >> n;
  int arr1[n], arr2[n];
  for (i = 0; i < n; i++) cin >> arr1[i];
  for (i = 0; i < n; i++) cin >> arr2[i];
  MergeSort(arr2, 0, n - 1);
  for (i = 0; i < n; i++) {
    sum = sum + arr1[i];
    if (arr2[n - 1] + arr2[n - 2] < sum) {
      break;
    }
  }
  if (sum <= arr2[n - 1] + arr2[n - 2])
    cout << "YES" << endl;
  else
    cout << "NO" << endl;
}

### Prompt

In CPP, your task is to solve 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;
void Merge(int *a, int low, int high, int mid) {
  int i, j, k, temp[high - low + 1];
  i = low;
  k = 0;
  j = mid + 1;
  while (i <= mid && j <= high) {
    if (a[i] < a[j]) {
      temp[k] = a[i];
      k++;
      i++;
    } else {
      temp[k] = a[j];
      k++;
      j++;
    }
  }
  while (i <= mid) {
    temp[k] = a[i];
    k++;
    i++;
  }
  while (j <= high) {
    temp[k] = a[j];
    k++;
    j++;
  }
  for (i = low; i <= high; i++) {
    a[i] = temp[i - low];
  }
}
void MergeSort(int *a, int low, int high) {
  int mid;
  if (low < high) {
    mid = (low + high) / 2;
    MergeSort(a, low, mid);
    MergeSort(a, mid + 1, high);
    Merge(a, low, high, mid);
  }
}
int main() {
  long int i, n, sum = 0;
  cin >> n;
  int arr1[n], arr2[n];
  for (i = 0; i < n; i++) cin >> arr1[i];
  for (i = 0; i < n; i++) cin >> arr2[i];
  MergeSort(arr2, 0, n - 1);
  for (i = 0; i < n; i++) {
    sum = sum + arr1[i];
    if (arr2[n - 1] + arr2[n - 2] < sum) {
      break;
    }
  }
  if (sum <= arr2[n - 1] + arr2[n - 2])
    cout << "YES" << endl;
  else
    cout << "NO" << endl;
}

```