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.*;
import java.util.*;

public class Main {
    static StreamTokenizer in=null;
    static List<Integer> dfs=new ArrayList<>();
    public static void main(String[] args) throws IOException {
        in = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
        PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
        InputReader reader = new InputReader(System.in);
        int t=nextInt();
        for(int i=0;i<t;i++){
            int n=nextInt();
            int []a=new int[n];
            int []b=new int[n];
            for(int j=0;j<n;j++)
                a[j]=nextInt();
            for(int j=0;j<n;j++)
                b[j]=nextInt();
            boolean c[]=new boolean[2];
            int j;
            for(j=0;j<n;j++){
                if(b[j]!=a[j]){
                    if(b[j]>a[j]&&!c[1]){
                            break;
                    }
                    if(b[j]<a[j]&&!c[0])
                            break;
                }

                if(a[j]==1)
                    c[1]=true;
                else if(a[j]==-1)
                    c[0]=true;
            }
            if(j==n)
                System.out.println("YES");
            else
                System.out.println("NO");
        }
    }
    static class InputReader{
        private BufferedReader in;
        private StringTokenizer tokenizer;
        public InputReader(InputStream stream){
            in = new BufferedReader(new InputStreamReader(stream),32768);
            tokenizer = null;
        }
        public String next() throws IOException{
            while(tokenizer==null || !tokenizer.hasMoreTokens()){
                tokenizer = new StringTokenizer(in.readLine());
            }
            return tokenizer.nextToken();
        }
        public Long nextLong() throws IOException {
            String next = next();
            return Long.valueOf(next);
        }
        public int nextInt() throws IOException{
            return Integer.valueOf(next());
        }
    }
    public static double nextDouble() throws IOException{ in.nextToken(); return in.nval; }
    public static float nextFloat() throws IOException{ in.nextToken(); return (float)in.nval; }
    public static int nextInt() throws IOException{ in.nextToken(); return (int)in.nval; }
    public static String next() throws IOException{ in.nextToken(); return in.sval; }//仅能读取纯字母字符串有数字会返回null
 }