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 _I() {
  int x;
  scanf("%d", &x);
  return x;
}
struct node {
  int l, h, r;
  node() {}
  node(int x, int y, int z) {
    l = x;
    h = y;
    r = z;
  }
};
void solve() {
  int n = _I();
  int ar[100001];
  long long sum = 0;
  for (int i = 0; i < n; i++) {
    long long x = _I();
    sum += x;
  }
  for (int i = 0; i < n; i++) {
    ar[i] = _I();
  }
  sort(ar, ar + n);
  if (ar[n - 1] + ar[n - 2] >= sum)
    printf("YES\n");
  else
    printf("NO\n");
}
int main() {
  solve();
  return 0;
}