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.*;
import java.util.*;

public class A {

	static int INF = (int) 1e9 + 1;

	public static void main(String[] args) throws IOException {
		Scanner sc = new Scanner();
		PrintWriter out = new PrintWriter(System.out);
		int tc = sc.nextInt();
		while (tc-- > 0) {
			int n = sc.nextInt(), a[] = sc.nxtArr(n), b[] = sc.nxtArr(n);
			int lo = 0, hi = Math.min(a[n - 1], b[n - 1]);
			int min = INF;
			while (lo <= hi) {
				int mid = lo + hi >> 1;
				boolean ok = true;
				int need = b[n - 1] - mid;
				for (int i = n - 1; i >= 0; i--) {
					int give = Math.min(b[i], need);
					need -= give;
					if (need > 0)
						ok = false;
					give = Math.min(a[i], b[i] - give);
					need = a[i] - give;

				}
				if (ok) {
					min = mid;
					hi = mid - 1;
				} else
					lo = mid + 1;

			}
			if (min == INF) {
				out.println("NO");
				continue;
			}
			a[n - 1] -= min;
			b[n - 1] -= min;
			a[0] -= (b[n - 1]);
			for (int i = n - 2; i >= 0; i--) {
				int give=Math.min(b[i], a[i+1]);
				a[i+1]-=give;
				give = Math.min(a[i], b[i] - give);
				a[i]-=give;

			}
			boolean ok=true;
			for(int x:a)
				ok&=x<=0;
			out.println(ok?"YES":"NO");
		}

		out.close();

	}

	static class Scanner {
		BufferedReader br;
		StringTokenizer st;

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

		Scanner(String fileName) throws FileNotFoundException {
			br = new BufferedReader(new FileReader(fileName));
		}

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

		String nextLine() throws IOException {
			return br.readLine();
		}

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

		long nextLong() throws NumberFormatException, IOException {
			return Long.parseLong(next());
		}

		double nextDouble() throws NumberFormatException, IOException {
			return Double.parseDouble(next());
		}

		boolean ready() throws IOException {
			return br.ready();
		}

		int[] nxtArr(int n) throws IOException {
			int[] ans = new int[n];
			for (int i = 0; i < n; i++)
				ans[i] = nextInt();
			return ans;
		}

	}

	static void sort(int[] a) {
		shuffle(a);
		Arrays.sort(a);
	}

	static void shuffle(int[] a) {
		int n = a.length;
		Random rand = new Random();
		for (int i = 0; i < n; i++) {
			int tmpIdx = rand.nextInt(n);
			int tmp = a[i];
			a[i] = a[tmpIdx];
			a[tmpIdx] = tmp;
		}
	}

}