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.util.*;
import java.io.*;
public class Edu90F {
   public static void main(String[] main) throws Exception{
      BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
      StringTokenizer st = new StringTokenizer(br.readLine());
      PrintWriter out = new PrintWriter(System.out);
      int t = Integer.parseInt(st.nextToken());
      for(int trial = 1; trial <= t; trial++){
         st = new StringTokenizer(br.readLine());
         int n = Integer.parseInt(st.nextToken());
         int[] a = new int[n];
         int[] b = new int[n];
         st = new StringTokenizer(br.readLine());
         for(int i = 0; i < n; i++) 
        	 a[i] = Integer.parseInt(st.nextToken());
         st = new StringTokenizer(br.readLine());
         for(int i = 0; i < n; i++) 
        	 b[i] = Integer.parseInt(st.nextToken());
         int min = Math.min(a[0], b[0]);
         if(a[0] >= b[0])
        	 a[0] -= b[0];
         else {
    		 b[0] -= a[0];
    		 a[0] = 0;
    		 if(a[1] >= b[0])
    			 a[1] -= b[0];
    		 else
    			 a[1] = 0;
    	 }
         boolean fail = false;
         for(int i = 1; i < n; i++) {
        	 if(a[i] > b[i] + min) {
        		 fail = true;
        		 break;
        	 }
        	 else if(a[i] > b[i]) {
        		 min -= a[i] - b[i];
        		 min = Math.min(min,b[i]);
        		 a[0] += a[i] - b[i];
        		 a[i] = 0;
        	 }
        	 else {
        		 min = Math.min(min,a[i]);
        		 b[i] -= a[i];
        		 a[i] = 0;
        		 if(a[(i+1)%n] >= b[i])
        			 a[(i+1)%n] -= b[i];
        		 else
        			 a[(i+1)%n] = 0;
        	 }
         }
         if(fail || a[0] > 0)
        	 out.println("NO");
         else
        	 out.println("YES");
      }
      out.close();
   }
}