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".

import java.io.*;
import java.util.*;
import java.math.*;
import static java.lang.Math.*;
import static java.lang.Integer.parseInt;
import static java.lang.Long.parseLong;
import static java.lang.Double.parseDouble;
import static java.lang.String.*;

public class Main {

    public static void main(String[] args) throws IOException {
     // BufferedReader in = new BufferedReader(new FileReader("cases.in"));
       BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
        //(new FileReader("input.in"));
        StringTokenizer tk;
        StringBuilder out = new StringBuilder();
        int n=parseInt(in.readLine());
        long []a=new long[n],b=new long[n];
        long sum=0;
        tk=new StringTokenizer(in.readLine());
        for (int i = 0; i < n; i++) {
            a[i]=parseLong(tk.nextToken());
            sum+=a[i];
        }
        tk=new StringTokenizer(in.readLine());
        for (int i = 0; i < n; i++) {
            b[i]=parseLong(tk.nextToken());
          
        }
      
        Arrays.sort(b);
         if((long)sum<=(long)b[n-1]+(long)b[n-2])
             System.out.println("YES");
         else System.out.println("NO");
    }
    
}
