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;
const long long maxn = 2e5 + 4;
const long long MAXN = 1000001;
long long power(long long x, unsigned int y) {
  long long temp;
  if (y == 0) return 1;
  temp = power(x, y / 2);
  if ((y % 2) == 0)
    return temp * temp;
  else
    return x * temp * temp;
}
vector<long long> a, b, c;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  long long n;
  cin >> n;
  long long o;
  for (long long i = 0; i < (long long)n; i++) cin >> o, a.push_back(o);
  for (long long i = 0; i < (long long)n; i++) cin >> o, b.push_back(o);
  c = b;
  sort(c.begin(), c.end());
  long long s = c[n - 1] + c[n - 2];
  long long t = 0;
  for (long long i = 0; i < (long long)n; i++) t += a[i];
  (t <= s) ? cout << "YES\n" : cout << "NO\n";
  return 0;
}