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.
/*
If you want to aim high, aim high
Don't let that studying and grades consume you
Just live life young
******************************
What do you think? What do you think?
1st on Billboard, what do you think of it
Next is a Grammy, what do you think of it
However you think, I’m sorry, but shit, I have no fcking interest
*******************************
I'm standing on top of my Monopoly board
That means I'm on top of my game and it don't stop
til my hip don't hop anymore
https://www.a2oj.com/Ladder16.html
*******************************
300IQ as writer = Sad!
*/
import java.util.*;
import java.io.*;
import java.math.*;

   public class x1373F
   {
      public static void main(String hi[]) throws Exception
      {
         BufferedReader infile = new BufferedReader(new InputStreamReader(System.in));  
         StringTokenizer st = new StringTokenizer(infile.readLine());
         int T = Integer.parseInt(st.nextToken());
         StringBuilder sb = new StringBuilder();
         matcha:while(T-->0)
         {
            int N = Integer.parseInt(infile.readLine());
            int[] arr = readArr(N, infile, st);
            int[] brr = readArr(N, infile, st);
            long temp = 0L;
            for(int x: arr)   temp += x;
            for(int x: brr)   temp -= x;
            if(temp > 0)
               sb.append("no\n");
            else
            {
               int low = 0;
               int high = brr[0];
               outer:while(low <= high)
               {
                  int mid = (low+high)/2;
                  long leftover = arr[0]-mid;
                  long carry = brr[0]-mid;
                  if(leftover < 0)
                  {
                     carry -= leftover;
                     leftover = 0;
                  }
                  for(int i=1; i < N; i++)
                  {
                     if(carry+brr[i] < arr[i])
                     {
                        high = mid-1;
                        continue outer;
                     }
                     carry = (carry+brr[i])-arr[i];
                     carry = Math.min(carry, brr[i]);
                  }
                  if(carry >= leftover)
                  {
                     sb.append("YES\n");
                     continue matcha;
                  }
                  else
                     low = mid+1;
               }
               sb.append("nO\n");
            }
         }
         System.out.print(sb);
      }
      public static int[] readArr(int N, BufferedReader infile, StringTokenizer st) throws Exception
      {
         int[] arr = new int[N];
         st = new StringTokenizer(infile.readLine());
         for(int i=0; i < N; i++)
            arr[i] = Integer.parseInt(st.nextToken());
         return arr;
      }
   }