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;
template <typename T>
inline void read(T &x) {
  short f = 1;
  x = 0;
  char c = getchar();
  while (!isdigit(c)) {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (isdigit(c)) x = (x << 3) + (x << 1) + (c ^ 48), c = getchar();
  x *= f;
}
template <typename T>
inline void write(T x) {
  if (x < 0) putchar('-'), x = -x;
  if (x > 9) write(x / 10);
  putchar(x % 10 + '0');
}
template <typename T>
inline void writeln(T x) {
  write(x), putchar('\n');
}
int main() {
  long long n, sum = 0, a[100010];
  read(n);
  for (int i = 1; i <= n; i++) {
    int x;
    read(x);
    sum += x;
  }
  for (int i = 1; i <= n; i++) read(a[i]);
  sort(a + 1, a + n + 1);
  if (a[n] + a[n - 1] >= sum)
    cout << "YES" << endl;
  else
    cout << "NO" << endl;
  return 0;
}
