The government of Berland decided to improve network coverage in his country. Berland has a unique structure: the capital in the center and n cities in a circle around the capital. The capital already has a good network coverage (so the government ignores it), but the i-th city contains a_i households that require a connection.

The government designed a plan to build n network stations between all pairs of neighboring cities which will maintain connections only for these cities. In other words, the i-th network station will provide service only for the i-th and the (i + 1)-th city (the n-th station is connected to the n-th and the 1-st city).

All network stations have capacities: the i-th station can provide the connection to at most b_i households.

Now the government asks you to check can the designed stations meet the needs of all cities or not — that is, is it possible to assign each household a network station so that each network station i provides the connection to at most b_i households.

Input

The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases.

The first line of each test case contains the single integer n (2 ≤ n ≤ 10^6) — the number of cities and stations.

The second line of each test case contains n integers (1 ≤ a_i ≤ 10^9) — the number of households in the i-th city.

The third line of each test case contains n integers (1 ≤ b_i ≤ 10^9) — the capacities of the designed stations.

It's guaranteed that the sum of n over test cases doesn't exceed 10^6.

Output

For each test case, print YES, if the designed stations can meet the needs of all cities, or NO otherwise (case insensitive).

Example

Input


5
3
2 3 4
3 3 3
3
3 3 3
2 3 4
4
2 3 4 5
3 7 2 2
4
4 5 2 3
2 3 2 7
2
1 1
10 10


Output


YES
YES
NO
YES
YES

Note

In the first test case: 

  * the first network station can provide 2 connections to the first city and 1 connection to the second city; 
  * the second station can provide 2 connections to the second city and 1 connection to the third city; 
  * the third station can provide 3 connections to the third city. 



In the second test case: 

  * the 1-st station can provide 2 connections to the 1-st city; 
  * the 2-nd station can provide 3 connections to the 2-nd city; 
  * the 3-rd station can provide 3 connections to the 3-rd city and 1 connection to the 1-st station. 



In the third test case, the fourth city needs 5 connections, but the third and the fourth station has 4 connections in total.
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Scanner;
import java.util.StringTokenizer;

public class netcov {
    static FastScanner in = new FastScanner();

    static void solve()
    {
        int n = in.nextInt();
        int[] a = new int[n];
        int[] b = new int[n];
        a = in.readArray(n);
        b = in.readArray(n);
        long need = 0;
        long add = 0;
        long extra = (long) 2e9;
        for(int i=n-1;i>=0;i--)
        {
            if(b[i] < need)
            {
                System.out.println("NO");
                return;
            }
            long val = b[i]-a[i];
            long to_add = Math.min(extra, Math.max(0, val-need));
            add += to_add;
            extra = Math.min(extra-to_add, b[i]-need-to_add);
            need = Math.max(0, need-val);
        }
        if(add >= need)
        {
            System.out.println("YES");
        }
        else
        {
            System.out.println("NO");
        }
    }

    public static void main(String[] args) {
        int tc = in.nextInt();
        for(int t=0;t<tc;t++)
        {
            solve();
        }
    }

    static class FastScanner
    {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer("");
        String next()
        {
            while(!st.hasMoreTokens())
            {
                try
                {
                    st = new StringTokenizer(br.readLine());
                }
                catch (IOException e)
                {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }
        int nextInt()
        {
            return Integer.parseInt(next());
        }
        int[] readArray(int n)
        {
            int[] a = new int[n];
            for(int i=0;i<n;i++) a[i] = nextInt();
            return a;
        }
        long nextLong()
        {
            return Long.parseLong(next());
        }
    }

}
