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".

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int a[100005], b[100005];
int n;
void Citire() {
  int i;
  cin >> n;
  for (i = 1; i <= n; i++) cin >> a[i];
  for (i = 1; i <= n; i++) cin >> b[i];
}
int Raspuns(long long x, long long y) {
  if (x < y) return 0;
  return 1;
}
void Rezolvare() {
  int i;
  long long m1, m2;
  long long c;
  m1 = m2 = c = 0LL;
  for (i = 1; i <= n; i++) {
    if (b[i] * 1LL > m1) m1 = 1LL * b[i];
    c += (1LL * a[i]);
  }
  for (i = 1; i <= n && b[i] != 1LL * m1; i++)
    ;
  b[i] = 0LL;
  for (i = 1; i <= n; i++)
    if (b[i] > 1LL * m2) m2 = b[i];
  m1 += m2;
  if (Raspuns(m1, c) == 1)
    cout << "YES";
  else
    cout << "NO";
}
int main() {
  Citire();
  Rezolvare();
  return 0;
}