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.StringTokenizer;

public class F {
	static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
	static StringTokenizer st;
	static PrintWriter out = new PrintWriter(System.out);
	private static int[][] cables;
	private static int[][] ciudades;
	private static final String Y = "YES";
	private static final String N = "NO";

	static void solve() throws Exception {
		int tests = nextInt();
		testcase: for (int t = 0; t < tests; t++) {
			int n = nextInt();
			ciudades = new int[n][2];
			cables = new int[n][3];
			for (int i = 0; i < n; i++) {
				ciudades[i] = new int[] { nextInt(), 0 };
			}
			for (int i = 0; i < n; i++) {
				cables[i] = new int[] { nextInt(), 0, 0 };
			}

			for (int i = 0; i < n; i++) {
				if (i > 0) {
					llenarIzq(i);
					if (!completed(i)) {// if izq no esta lleno
						out.println(N);
						continue testcase;
					}
				}
				llenarDerecha(i);
			}

			for (int i = 0; i < n; i++) {

				if (completed(i)) {
					out.println(Y);
					continue testcase;
				} else {// izq no esta llena
					llenarIzq(i);
					if (completed(i)) {
						out.println(Y);
						continue testcase;
					} else { // izq no esta llena
						completarIzq(i);
						if (!completed(i)) { // aun asi no se puede llenar la izq
							out.println(N);
							continue testcase;
						}
					}
				}
			}
			if (completed(0)) {
				out.println(Y);
			} else {
				out.println(N);
			}
//			String r = Y;
//			for (int i = 0; i < n; i++) {
//				if (!completed(i)) {
//					r = N;
//					break;
//				}
//			}
//			out.println(r);
		}
	}

	private static int disponibles(int i) {
		return cables[i][0] - cables[i][1] - cables[i][2];
	}

	private static void llenarDerecha(int i) {
		int pongo = Math.min(disponibles(i), faltantes(nexti(i)));
		cables[i][2] += pongo;
		ciudades[nexti(i)][1] += pongo;
	}

	private static int nexti(int i) {
		return (i + 1) % ciudades.length;
	}

	private static void llenarIzq(int i) {
		int pongo = Math.min(disponibles(i), faltantes(i));
		cables[i][1] += pongo;
		ciudades[i][1] += pongo;
	}

	private static void completarIzq(int i) {
		int sacoypongo = Math.min(faltantes(i), cables[i][2]);
		cables[i][2] -= sacoypongo;
		cables[i][1] += sacoypongo;
		ciudades[nexti(i)][1] -= sacoypongo;
		ciudades[i][1] += sacoypongo;
	}

	private static boolean completed(int i) {
		return ciudades[i][1] == ciudades[i][0];
	}

	private static int faltantes(int i) {
		return ciudades[i][0] - ciudades[i][1];
	}

	public static void main(String[] args) throws Exception {
		solve();
		out.close();
	}

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

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

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