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.*;

public class Main {
    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        int n = Integer.parseInt(br.readLine());
        String[] s = br.readLine().split("\\s");
        int[] volumes = new int[n];
        for(int i=0;i<n;++i) volumes[i] = Integer.parseInt(s[i]);
        s = br.readLine().split("\\s");
        int[] capacity = new int[n];
        for(int i=0;i<n;++i) capacity[i] = Integer.parseInt(s[i]);
        System.out.println(solve(n,volumes,capacity));   
    }
    
    private static String solve(int n,int[] volumes,int[] capacity){
        long sum = 0;
        for(int i=0;i<volumes.length;++i){
            sum += volumes[i];
        }
        
        int max = 0;
        for(int i=1;i<capacity.length;++i){
            if(capacity[max] < capacity[i]) max = i;
        }
        
        int second_max = -1;
        for(int i=0;i<capacity.length;++i){
            if(i == max) continue;
            if(capacity[i] <= capacity[max] && (second_max == -1 || capacity[second_max] < capacity[i])){
                second_max = i;
            }
        }
        
        if(sum > (long)capacity[max] + (long)capacity[second_max]) return "NO";
        return "YES";
    }
}