Once again, Boris needs the help of Anton in creating a task. This time Anton needs to solve the following problem:

There are two arrays of integers a and b of length n. It turned out that array a contains only elements from the set \{-1, 0, 1\}.

Anton can perform the following sequence of operations any number of times:

  1. Choose any pair of indexes (i, j) such that 1 ≤ i < j ≤ n. It is possible to choose the same pair (i, j) more than once. 
  2. Add a_i to a_j. In other words, j-th element of the array becomes equal to a_i + a_j. 



For example, if you are given array [1, -1, 0], you can transform it only to [1, -1, -1], [1, 0, 0] and [1, -1, 1] by one operation.

Anton wants to predict if it is possible to apply some number (zero or more) of these operations to the array a so that it becomes equal to array b. Can you help him?

Input

Each test contains multiple test cases. 

The first line contains the number of test cases t (1 ≤ t ≤ 10000). The description of the test cases follows.

The first line of each test case contains a single integer n (1 ≤ n ≤ 10^5) — the length of arrays.

The second line of each test case contains n integers a_1, a_2, ..., a_n (-1 ≤ a_i ≤ 1) — elements of array a. There can be duplicates among elements.

The third line of each test case contains n integers b_1, b_2, ..., b_n (-10^9 ≤ b_i ≤ 10^9) — elements of array b. There can be duplicates among elements.

It is guaranteed that the sum of n over all test cases doesn't exceed 10^5.

Output

For each test case, output one line containing "YES" if it's possible to make arrays a and b equal by performing the described operations, or "NO" if it's impossible.

You can print each letter in any case (upper or lower).

Example

Input


5
3
1 -1 0
1 1 -2
3
0 1 1
0 2 2
2
1 0
1 41
2
-1 0
-1 -41
5
0 1 -1 1 -1
1 1 -1 1 -1


Output


YES
NO
YES
YES
NO

Note

In the first test-case we can choose (i, j)=(2, 3) twice and after that choose (i, j)=(1, 2) twice too. These operations will transform [1, -1, 0] → [1, -1, -2] → [1, 1, -2]

In the second test case we can't make equal numbers on the second position.

In the third test case we can choose (i, j)=(1, 2) 41 times. The same about the fourth test case.

In the last lest case, it is impossible to make array a equal to the array b.
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.InputMismatchException;

public class ques2 {

	static class Reader {

		private InputStream mIs;
		private byte[] buf = new byte[1024];
		private int curChar;
		private int numChars;

		public Reader() {
			this(System.in);
		}

		public Reader(InputStream is) {
			mIs = is;
		}

		public int read() {
			if (numChars == -1) {
				throw new InputMismatchException();

			}
			if (curChar >= numChars) {
				curChar = 0;
				try {
					numChars = mIs.read(buf);
				} catch (IOException e) {
					throw new InputMismatchException();
				}
				if (numChars <= 0) {
					return -1;
				}
			}
			return buf[curChar++];
		}

		public String nextLine() {
			int c = read();
			while (isSpaceChar(c)) {
				c = read();
			}
			StringBuilder res = new StringBuilder();
			do {
				res.appendCodePoint(c);
				c = read();
			} while (!isEndOfLine(c));
			return res.toString();
		}

		public String next() {
			int c = read();
			while (isSpaceChar(c)) {
				c = read();
			}
			StringBuilder res = new StringBuilder();
			do {
				res.appendCodePoint(c);
				c = read();
			} while (!isSpaceChar(c));
			return res.toString();
		}

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

		public long nextLong() {
			int c = read();
			while (isSpaceChar(c)) {
				c = read();
			}
			int sgn = 1;
			if (c == '-') {
				sgn = -1;
				c = read();
			}
			long res = 0;
			do {
				if (c < '0' || c > '9') {
					throw new InputMismatchException();
				}
				res *= 10;
				res += c - '0';
				c = read();
			} while (!isSpaceChar(c));
			return res * sgn;
		}

		public int nextInt() {
			int c = read();
			while (isSpaceChar(c)) {
				c = read();
			}
			int sgn = 1;
			if (c == '-') {
				sgn = -1;
				c = read();
			}
			int res = 0;
			do {
				if (c < '0' || c > '9') {
					throw new InputMismatchException();
				}
				res *= 10;
				res += c - '0';
				c = read();
			} while (!isSpaceChar(c));
			return res * sgn;
		}

		public int[] intArray(int n) {
			Reader r = new Reader();
			int arr[] = new int[n];
			for (int i = 0; i < n; i++) {
				arr[i] = r.nextInt();
			}
			return arr;
		}

		public long[] longArray(int n) {
			Reader r = new Reader();
			long arr[] = new long[n];
			for (int i = 0; i < n; i++) {
				arr[i] = r.nextInt();
			}
			return arr;
		}

		public void printArray(int arr[]) {
			PrintWriter out = new PrintWriter(System.out);
			for (int i = 0; i < arr.length; i++) {
				if (i == arr.length - 1)
					out.println(arr[i]);
				else
					out.print(arr[i] + " ");
			}
			out.close();
		}

		public boolean isSpaceChar(int c) {
			return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
		}

		public boolean isEndOfLine(int c) {
			return c == '\n' || c == '\r' || c == -1;
		}
	}

	public static void main(String[] args) throws IOException {

		Reader r = new Reader();

		int t = r.nextInt();
		StringBuilder res = new StringBuilder();

		while (t-- > 0) {

			int n = r.nextInt();

			int a[] = new int[n];
			for (int i = 0; i < n; i++) {
				a[i] = r.nextInt();
			}

			int b[] = new int[n];
			for (int i = 0; i < n; i++) {
				b[i] = r.nextInt();
			}

			boolean flag = true;
			if (a[0] != b[0]) {
				flag = false;
			} else {
				boolean neg = false, pos = false;

				if (a[0] > 0) {
					pos = true;
				} else if (a[0] < 0) {
					neg = true;
				}

				for (int i = 1; i < n; i++) {
					if (b[i] > a[i] && !pos) {
						flag = false;
						break;
					} else if (b[i] < a[i] && !neg) {
						flag = false;
						break;
					} else {
						if (a[i] > 0) {
							pos = true;
						}
						if (a[i] < 0) {
							neg = true;
						}
					}
				}
			}
			if (flag)
				res.append("YES");
			else
				res.append("NO");
			res.append("\n");
		}

		System.out.print(res);

	}

}