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 int maxn = 100005;
long long a[maxn];
long long b[maxn];
int main() {
  int n;
  while (~scanf("%d", &n)) {
    priority_queue<long long> pq;
    while (!pq.empty()) pq.pop();
    long long suma = 0, p = 0, q = 0;
    for (int i = 0; i < n; i++) {
      scanf("%lld", &a[i]);
      suma += a[i];
    }
    for (int i = 0; i < n; i++) {
      scanf("%lld", &b[i]);
      pq.push(b[i]);
    }
    p = pq.top();
    pq.pop();
    q = pq.top();
    if (p + q >= suma)
      puts("YES");
    else
      puts("NO");
  }
  return 0;
}