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.util.Scanner;

/*
 * To change this license header, choose License Headers in Project Properties.
 * To change this template file, choose Tools | Templates
 * and open the template in the editor.
 */
/**
 *
 * @author Kayali
 */
public class JavaApplication5 {

    /**
     * @param args the command line arguments
     */
    public static void main(String[] args) {
        int n;
                double s = 0;
        Scanner in=new Scanner(System.in);
           n = in.nextInt();
           if (n==2)
           {System.out.println("YES");
           }
           else{
           double[] a = new double[n];
           double[] b = new double[n];
            
         for (int i = 0; i < n; i++)
           {
              
              a[i]  = in.nextDouble();
                 s+=a[i];    
            }
           for (int i = 0; i < n; i++)
            {
             
               b[i] = in.nextDouble();

            }          
            double max = b[0];
                   int l=0;
            for (int i = 1; i < n; i++)
            {if (max < b[i])
                {max = b[i];
                    l = i;}}
            
           b[l]=0;
            double max1 =0;
            for (int i = 1; i < n; i++)
            {
                if (max1 < b[i])
                {
                    max1 = b[i];
                    
                }
            }
            if((max+max1)>=s)
            System.out.println("YES");
            else System.out.println("NO");
          
           }
    }
    
}
