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.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.HashMap;
import java.util.HashSet;
import java.util.LinkedList;
import java.util.Scanner;
import java.util.Set;
import java.util.Stack;
import java.util.StringTokenizer;
import java.util.TreeSet;

public class code6 {

	static class FastReader {
		BufferedReader br;
		StringTokenizer st;

		public FastReader() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}

		String next() {
			while (st == null || !st.hasMoreElements()) {
				try {
					st = new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			}
			return st.nextToken();
		}

		int nextInt() {
			return Integer.parseInt(next());
		}

		long nextLong() {
			return Long.parseLong(next());
		}

		double nextDouble() {
			return Double.parseDouble(next());
		}

		String nextLine() {
			String str = "";
			try {
				str = br.readLine();
			} catch (IOException e) {
				e.printStackTrace();
			}
			return str;
		}
	}
	public static void main(String[] args) throws CloneNotSupportedException {
		// TODO Auto-generated method stub
		FastReader scn = new FastReader();
		PrintWriter out = new PrintWriter(System.out);
		int t = scn.nextInt();
		//int p = t;
		while(t-->0) {
			int n = scn.nextInt();
			int[] a = new int[n];
			int[] b = new int[n];
			for(int i=0; i<n; i++)
				a[i] = scn.nextInt();
			for(int i=0; i<n; i++)
				b[i] = scn.nextInt();
			int l = 0;
			int hi = b[0];
			boolean ans = false;
			while(l<=hi) {
				int mid = (l+hi)/2;
				int p = mid;
				int flag = 0;
				for(int i=1; i<n; i++) {
					if(b[i]+p<a[i]) {
						flag = 1;
						break;
					}
					p = b[i]+p-a[i];
					p = Math.min(b[i], p);
				}
				if(flag == 1)
					l = mid+1;
				else {
					if(b[0]-mid+p<a[0])
						hi = mid-1;
					else {
						ans = true;
						//System.out.println(mid);
						break;
					}
				}
			}
			
			if(ans)
				System.out.println("YES");
			else
				System.out.println("NO");
		}
		out.flush();
	}
	
  
	public static boolean check(int mid, int c, int[] arr, int k) {
		int ans = 0;
		for(int i=0; i<arr.length; i++) {
			if(c==0) {
				ans++;
				c = (c+1)%2;
			}else {
				if(arr[i]<=mid) {
					ans++;
					c = (c+1)%2;
				}
			}
		}
		return ans>=k;
	}
	public static int power(int x, int y, int p) 
    { 
        // Initialize result 
        int res = 1;      
         
        // Update x if it is more   
        // than or equal to p 
        x = x % p;  
  
       if (x == 0) return 0; // In case x is divisible by p;    
  
        while (y > 0) 
        { 
            // If y is odd, multiply x 
            // with result 
            if((y & 1)==1) 
                res = (res * x) % p; 
      
            // y must be even now 
            // y = y / 2 
            y = y >> 1;  
            x = (x * x) % p;  
        } 
        return res; 
    } 
	
	public static class Pair implements Comparable<Pair>{
		int m;
		int s;
		Pair(int m, int s){
			this.m = m;
			this.s = s; 
		}
		public int compareTo(Pair p) {
			return this.m-p.m;
		}
	}
    
}

