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".
//author: Ala Abid
import java.io.*;
import java.util.*;

public class NewMain {
    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer(br.readLine());
        int n=Integer.parseInt(st.nextToken());
        int[] rem = new int[n], cap = new int[n];
        long tot=0,cap1=0,cap2=0;
        st = new StringTokenizer(br.readLine());
        for(int i=0;i<n;i++){
            rem[i]=Integer.parseInt(st.nextToken());
            tot+=rem[i];
        }
        st = new StringTokenizer(br.readLine());
        for(int i=0;i<n;i++){
            cap[i]=Integer.parseInt(st.nextToken());
            if(cap[i]>cap1) {
                cap2=cap1;
                cap1=cap[i];
            }
            else if(cap[i]>cap2)cap2=cap[i];
        }
        if(cap1+cap2-tot<0)System.out.println("NO");
        else System.out.println("YES");
    }
}
