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.*;

public class Greed{
    public static void main(String[] args){
        int n;
        long sum = 0;
        Scanner sc = new Scanner(System.in);
        n = sc.nextInt();

        int[] rem = new int[n];
        int[] cap = new int[n+5];

        for(int i = 0; i < n; i++){
            rem[i] = sc.nextInt();
            sum += rem[i];
        }
        int max1 = n, max2 = n+1;
        cap[max1] = -1; cap[max2] = -1;

        for(int i = 0; i < n; i++){
            cap[i] = sc.nextInt();
            if(cap[i] > cap[max1]){
                max1 = i;   //ith index holds max value
            }
        }

        
        for(int i = 0; i < n; i++){

            if(i == max1){
                continue;
            }
            if(cap[i] > cap[max2]){
                max2 = i;
            }
        }
        
        int tot = cap[max1] + cap[max2];

        if(sum > tot){
            System.out.println("NO");
        }
        else{
            System.out.println("YES");
        }
    }
}