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 unsigned int mod = 1e7 + 7;
const long long int N = 1e5;
long long int lazy[400009], segtree[400009], x, y, n, d, color[200009], cnt;
bool visit[100009] = {false};
vector<long long int> v[100009];
int main() {
  long long int ans = 0, l, n, k = 0, q = 0, p = 0, t = 0, z = 0, w = INT_MAX,
                pre = 1e5, mid, a[100009] = {0}, h[26] = {0}, b[26] = {0}, t1,
                t2, now, r = 0, i = 0, j = 0, flag = 0, temp = 0, e = -1e4,
                o = 0;
  cin >> n;
  for (i = 0; i < n; i++) {
    cin >> l;
    k = k + l;
  }
  for (i = 0; i < n; i++) {
    cin >> a[i];
  }
  sort(a, a + n);
  if (a[n - 1] + a[n - 2] >= k)
    cout << "YES";
  else
    cout << "NO";
  return 0;
}