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;
long long expo(long long x, long long n, long long M) {
  if (n == 0)
    return 1;
  else if (n % 2 == 0)
    return (expo((x * x) % M, n / 2, M)) % M;
  else
    return (x * expo((x * x) % M, (n - 1) / 2, M)) % M;
}
bool sortbysec(pair<pair<long long, long long>, long long> p1,
               pair<pair<long long, long long>, long long> p2) {
  if (p1.first.first == p2.first.first)
    return p1.first.second > p2.first.second;
  return p1.first.first < p2.first.first;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long i, j, k, a, b, c, t, x, y, n, m, z, flag = 0, ans = 0, sum = 0;
  t = 1;
  while (t--) {
    cin >> n;
    vector<long long> v(n), v1(n);
    for (i = 0; i < n; i++) {
      cin >> v[i];
      sum += v[i];
    }
    for (i = 0; i < n; i++) {
      cin >> v1[i];
    }
    sort(v1.rbegin(), v1.rend());
    if (v1[0] + v1[1] >= sum) {
      cout << "YES";
    } else {
      cout << "NO";
    }
  }
  return 0;
}
