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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Arrays;
import java.util.stream.IntStream;

public class A {
    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        int n = Integer.parseInt(br.readLine());
        int[] a = new int[n];
        int[] b = new int[n];
        String[] in;
        in = br.readLine().split("\\s");
        for (int i = 0; i < n; i++)
            a[i] = Integer.parseInt(in[i]);
        in = br.readLine().split("\\s");
        for (int i = 0; i < n; i++)
            b[i] = Integer.parseInt(in[i]);
        long sum = sum(a);
        sum -= max(b);
        sum -= max(b);
        System.out.println(sum <= 0 ? "YES" : "NO");
    }

    private static long sum(int[] a) {
        long sum = 0;
        for (int i : a)
            sum += i;
        return sum;
    }

    private static int max(int[] b) {
        int max = Integer.MIN_VALUE;
        int ii = 0;
        for (int i = 0; i < b.length; i++)
            if (b[i] > max) {
            max = b[i];
            ii = i;
            }
        b[ii] = 0;
        return max;
    }
}