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.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.Queue;
import java.util.StringTokenizer;
import java.util.*;

public class a {
	static long mod = 1000000009L;

	public static void main(String[] args) throws Exception {
		in = new BufferedReader(new InputStreamReader(System.in));
		out = new PrintWriter(System.out);
		sc = new FastScanner(in);
		int t = sc.nextInt();
		while (t-- > 0) {
			int n = sc.nextInt();
			long[] a = new long[n];
			long[] b = new long[n];
			for (int i = 0; i < n; i++) {
				a[i] = sc.nextLong();
			}
			for (int i = 0; i < n; i++) {
				b[i] = sc.nextLong();
			}
			long lo = 0;
			long hi = b[0];
			boolean ch = false;
			while (lo <= hi) {
				long mid = (lo + hi) / 2;
				int res =pos(a, b, mid);
				if (res ==2 ) {
					ch = true;
					break;
				}
				if( ( b[n-1] < a[0] - mid ) ||( res==4 && mid<a[0])  ) {
					lo = mid +1;

				}else {
					hi = mid-1 ;
				}
			}
			if(ch) {
				out.println("YES");
			}else {
				out.println("NO");
			}
		}
		out.flush();
	}

	private static int pos(long[] a, long[] b, long mid) {
		int n = a.length;
		if(b[n-1] < a[0] - mid) {
			return 1;
		}
		long have = b[0] - mid ;
		for (int i = 1; i < n-1; i++) {
			if(a[i] > have + b[i]) {
				return 3;
			}
			long rem = Math.max(a[i] - have , 0 );
			
			have = b[i] - rem;
			
		}
		if(a[n-1] > have + b[n-1]-(  Math.max( a[0] - mid,0   )))return 4;
		
		return 2;
	}

	static BufferedReader in;
	static FastScanner sc;
	static PrintWriter out;

	static class FastScanner {
		BufferedReader in;
		StringTokenizer st;

		public FastScanner(BufferedReader in) {
			this.in = in;
		}

		public String nextToken() throws Exception {
			while (st == null || !st.hasMoreTokens()) {
				st = new StringTokenizer(in.readLine());
			}
			return st.nextToken();
		}

		public int nextInt() throws Exception {
			return Integer.parseInt(nextToken());
		}

		public long nextLong() throws Exception {
			return Long.parseLong(nextToken());
		}

		public double nextDouble() throws Exception {
			return Double.parseDouble(nextToken());
		}
	}
}