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 B {
	public static void main(String[] args) throws Exception {
		// TODO Auto-generated method stub
		InputReader s = new InputReader(System.in);
		PrintWriter p = new PrintWriter(System.out);
		int t = s.nextInt();
		while (t-- > 0) {
			int n = s.nextInt();
			long[] a = new long[n];
			long[] b = new long[n];
			int[][] cal = new int[n][2];
			int valpos = 0, valneg = 0;
			for (int i = 0; i < n; i++) {
				a[i] = s.nextInt();
				cal[i][0] = valpos;
				cal[i][1] = valneg;
				if (a[i] == 1)
					valpos = 1;
				else if (a[i] == -1)
					valneg = 1;
			}
			for (int i = 0; i < n; i++) {
				b[i] = s.nextInt();
			}
			int flag = 0;
			for (int i = 0; i < n; i++) {
				long val1 = a[i];
				long val2 = b[i];
				if (val1 == val2)
					continue;
				else if (val1 < val2) {
					if (cal[i][0] == 0) {
						flag = 1;
						break;
					}
				} else if (val1 > val2) {
					if (cal[i][1] == 0) {
						flag = 1;
						break;
					}
				}
			}
			if (flag == 0)
				p.println("YES");
			else
				p.println("NO");
		}
		p.flush();
		p.close();
	}

	public static long GCD(long a, long b) {
		if (b == 0)
			return a;
		else
			return GCD(b, a % b);
	}

	public static long LCM(long a, long b) {
		return (a * b) / GCD(a, b);
	}

	static class pair implements Comparable<pair> {
		Integer x, y;

		pair(int x, int y) {
			this.x = x;
			this.y = y;
		}

		public int compareTo(pair o) {
			int result = x.compareTo(o.x);
			if (result == 0)
				result = y.compareTo(o.y);

			return result;
		}

		public String toString() {
			return x + " " + y;
		}

		public boolean equals(Object o) {
			if (o instanceof pair) {
				pair p = (pair) o;
				return p.x - x == 0 && p.y - y == 0;
			}
			return false;
		}

		public int hashCode() {
			return new Long(x).hashCode() * 31 + new Long(y).hashCode();
		}
	}

	static class InputReader {

		private final InputStream stream;
		private final byte[] buf = new byte[8192];
		private int curChar, snumChars;
		private SpaceCharFilter filter;

		public InputReader(InputStream stream) {
			this.stream = stream;
		}

		public int snext() {
			if (snumChars == -1)
				throw new InputMismatchException();
			if (curChar >= snumChars) {
				curChar = 0;
				try {
					snumChars = stream.read(buf);
				} catch (IOException e) {
					throw new InputMismatchException();
				}
				if (snumChars <= 0)
					return -1;
			}
			return buf[curChar++];
		}

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

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

		public int[] nextIntArray(int n) {
			int a[] = new int[n];
			for (int i = 0; i < n; i++) {
				a[i] = nextInt();
			}
			return a;
		}

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

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

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

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

		public interface SpaceCharFilter {
			public boolean isSpaceChar(int ch);
		}
	}

	static class CodeX {
		public static void sort(long arr[]) {
			merge_sort(arr, 0, arr.length - 1);
		}

		private static void merge_sort(long A[], long start, long end) {
			if (start < end) {
				long mid = (start + end) / 2;
				merge_sort(A, start, mid);
				merge_sort(A, mid + 1, end);
				merge(A, start, mid, end);
			}

		}

		private static void merge(long A[], long start, long mid, long end) {
			long p = start, q = mid + 1;
			long Arr[] = new long[(int) (end - start + 1)];
			long k = 0;

			for (int i = (int) start; i <= end; i++) {
				if (p > mid)
					Arr[(int) k++] = A[(int) q++];

				else if (q > end)
					Arr[(int) k++] = A[(int) p++];

				else if (A[(int) p] < A[(int) q])
					Arr[(int) k++] = A[(int) p++];

				else
					Arr[(int) k++] = A[(int) q++];
			}
			for (int i = 0; i < k; i++) {
				A[(int) start++] = Arr[i];
			}

		}

	}

}