Problem: 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".

Solution (in JAVA):

Solution:
import java.io.*;
import java.lang.reflect.Array;
import java.util.*;
public class Main
{
static BufferedReader in=new BufferedReader(new InputStreamReader(System.in));
static StringTokenizer tok;
static boolean hasNext()
{
while(tok==null||!tok.hasMoreTokens())
try{
tok=new StringTokenizer(in.readLine());}
catch(Exception e){
return false;
}
return true;}
static String next()
{
hasNext();
return tok.nextToken();}
static long nextLong()
{
return Long.parseLong(next());}
static int nextInt()
{return Integer.parseInt(next());}
static PrintWriter out=new PrintWriter(new OutputStreamWriter(System.out));
static int n,m,n1,m1;
	public static void main(String[] args) {
		n=nextInt();
		long sum=0;
		int b[]=new int[n];
		for(int i=0;i<n;i++) {
			m=nextInt();
			sum+=m;
		}
		for(int i=0;i<n;i++) {
			b[i]=nextInt();
		}
		Arrays.sort(b);
		if(b[n-1]+b[n-2]>=sum)out.print("YES");
		else out.print("NO");
		out.flush();
	}
}