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 CF1373F extends PrintWriter {
	CF1373F() { super(System.out); }
	static class Scanner {
		Scanner(InputStream in) { this.in = in; } InputStream in;
		int k, l; byte[] bb = new byte[1 << 15];
		byte getc() {
			if (k >= l) {
				k = 0;
				try { l = in.read(bb); } catch (IOException e) { l = 0; }
				if (l <= 0) return -1;
			}
			return bb[k++];
		}
		int nextInt() {
			byte c = 0; while (c <= 32) c = getc();
			int a = 0; while (c > 32) { a = a * 10 + c - '0'; c = getc(); }
			return a;
		}
	}
	Scanner sc = new Scanner(System.in);
	public static void main(String[] $) {
		CF1373F o = new CF1373F(); o.main(); o.flush();
	}

	void main() {
		int t = sc.nextInt();
		while (t-- > 0) {
			int n = sc.nextInt();
			int[] aa = new int[n];
			int[] bb = new int[n];
			long sum = 0;
			for (int i = 0; i < n; i++)
				sum += aa[n - 1 - i] = sc.nextInt();
			for (int i = 0; i < n; i++)
				sum -= bb[n - 1 - i] = sc.nextInt();
			if (sum > 0) {
				println("NO");
				continue;
			}
			int[] dd = new int[n * 2];
			for (int i = 0; i < n; i++)
				dd[i] = dd[n + i] = aa[i] - bb[i];
			long[] ss = new long[n * 2];
			int[] ii = new int[n * 2]; int head = 0, cnt = 0;
			sum = 0;
			boolean yes = true;
			for (int i = 0; i < n * 2; i++) {
				sum += dd[i];
				long s = dd[i] - sum;
				while (cnt > 0 && ss[head + cnt - 1] <= s)
					cnt--;
				ss[head + cnt] = s;
				ii[head + cnt] = i;
				cnt++;
				if (ss[head] + sum > bb[(i + 1) % n]) {
					yes = false;
					break;
				}
				if (ii[head] == i - (n - 2)) {
					head++; cnt--;
				}
			}
			println(yes ? "YES" : "NO");
		}
	}
}
