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.BufferedReader;
    import java.io.IOException;
    import java.io.InputStreamReader;
    import java.util.StringTokenizer;
    import java.io.PrintWriter;
    import java.util.ArrayList;
    import java.util.List;
    import java.util.HashMap;
    import java.util.TreeMap;
    import java.util.Map;
    import java.util.Arrays;
    import java.util.Set;
    import java.util.HashSet;
    import java.util.Iterator;
    import java.util.Collections;
    import java.util.LinkedList;
    import java.util.Stack;
    
    /*
     *
     *  @author Riddle
     *
     */

    public final class KindAnton
    {
        final static long MOD = (long)1e9+7;

        static class Pair
        {
            int first;
            int second;
            public Pair(int first, int second)
            {
                this.first = first;
                this.second = second;
            }

            @Override
            public boolean equals(Object obj)
            {
                Pair p = (Pair)obj;
                return p.first==first && p.second==second;
            }
        }
        
        static class FastReader
        {
            BufferedReader br;
            StringTokenizer st;
     
            public FastReader()
            {
                br = new BufferedReader(new
                         InputStreamReader(System.in));
            }
     
            String next()
            {
                while (st == null || !st.hasMoreElements())
                {
                    try
                    {
                        st = new StringTokenizer(br.readLine());
                    }
                    catch (IOException  e)
                    {
                        e.printStackTrace();
                    }
                }
                return st.nextToken();
            }
     
            int nextInt()
            {
                return Integer.parseInt(next());
            }
     
            long nextLong()
            {
                return Long.parseLong(next());
            }
     
            double nextDouble()
            {
                return Double.parseDouble(next());
            }
     
            String nextLine()
            {
                String str = "";
                try
                {
                    str = br.readLine();
                }
                catch (IOException e)
                { 
                    e.printStackTrace();
                }
                return str;
            }

            int[] readArray(int arr[], int n)
            {
                for(int i=0;i<n;i++)
                {
                    arr[i] = nextInt();
                }

                return arr;
            }   
        }

        public static void main(String args[])
        {
            FastReader reader = new FastReader();
            PrintWriter out = new PrintWriter(System.out);
            int test = reader.nextInt();
            for(int t =0;t<test;t++)
            {
                int n = reader.nextInt();
                int a[] =  new int[n];
                int b[] = new int[n];
                a = reader.readArray(a, n);
                b = reader.readArray(b, n);

                if(a[0]!=b[0])
                {
                    out.println("NO");
                    continue;
                }
                int x = Integer.MAX_VALUE;
                int y = Integer.MAX_VALUE;
                for(int i=0;i<n;i++)
                {
                    if(a[i]==1&&x==Integer.MAX_VALUE)
                    {
                        x = i;
                    }
                    if(a[i]==-1&&y==Integer.MAX_VALUE)
                    {
                        y = i;
                    }
                }
                boolean result = true;
                for(int i=n-1;i>0;i--)
                {
                    if(a[i] == b[i])
                    {
                        continue;
                    }
                    if(a[i]>b[i])
                    {
                        if(y>=i)
                        {
                            result = false;
                            break;
                        }
                    }

                    if(a[i]<b[i])
                    {
                        if(x>=i)
                        {
                            result = false;
                            break;
                        }
                    }
                }

                String ans = result ? "YES" : "NO";
                out.println(ans);
            }
            out.flush();        
            out.close();
        }

    }