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;

public class Main30 {

    public static void main(String []args){
        Scanner scanner = new Scanner(System.in);
        int n = scanner.nextInt();
        long []a = new long[n];
        long []b = new long[n];

        long problem = 0, available = 0;
        for(int i=0;i<n;i++){
            a[i] = scanner.nextLong();
            problem += a[i];
        }

        for(int i=0;i<n;i++){
            b[i] = scanner.nextLong();
            //available += b[i]-a[i];
        }

        long maxiEffect1 = -1, maxiEffect2 = -1;
        int ind1 = 0, ind2 = 1;
        for(int i=0;i<n;i++){
            if(a[i] + (b[i]-a[i]) > maxiEffect1){
                maxiEffect2 = maxiEffect1;
                ind2 = ind1;
                maxiEffect1 = a[i] + (b[i]-a[i]);
                ind1 = i;
            }else if(a[i] + (b[i]-a[i]) > maxiEffect2){
                maxiEffect2 = a[i] + (b[i]-a[i]);
                ind2 = i;
            }
        }

        problem -= a[ind1];
        problem -= a[ind2];
        available += b[ind1]-a[ind1];
        available += b[ind2]-a[ind2];
        if(problem <= available){
            System.out.print("YES");
        }else{
            System.out.print("NO");
        }
    }
}
