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.
#include <bits/stdc++.h>
using namespace std;
long long int mod = 1000000007;
long long fast_exp(long long a, long long n, long long mod1) {
  if (n == 0)
    return 1;
  else if (n == 1)
    return a;
  if (n % 2 == 0)
    return fast_exp((a * a) % mod1, n / 2, mod1);
  else
    return (a * fast_exp((a * a) % mod1, n / 2, mod1)) % mod1;
}
static bool is_power[1000001] = {false};
void sieve() {
  for (long long int c = 2; c <= 1000000; c++)
    if (is_power[c] == true)
      continue;
    else
      for (long long int c1 = c * c; c1 <= 1000000; c1 = c1 * c)
        is_power[c1] = true;
  return;
}
int main() {
  sieve();
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  cin >> t;
  for (; t > 0; t--) {
    int n;
    cin >> n;
    vector<int> arr(n, 0), arr2(n, 0);
    for (int c = 0; c < n; c++) cin >> arr[c];
    for (int c1 = 0; c1 < n; c1++) cin >> arr2[c1];
    int is_one = 0;
    int is_min = 0;
    if (arr[0] != arr2[0])
      cout << "NO" << endl;
    else {
      for (int c = 0; c < n; c++)
        if (arr[c] == 1)
          is_one++;
        else if (arr[c] == -1)
          is_min++;
      int flag = 0;
      for (int c = n - 1; c >= 0; c--) {
        if (arr[c] == 1)
          is_one--;
        else if (arr[c] == -1)
          is_min--;
        if ((arr[c] > arr2[c] && is_min == 0) ||
            (arr[c] < arr2[c] && is_one == 0)) {
          cout << "NO" << endl, flag = -1;
          break;
        }
      }
      if (flag == 0) cout << "YES" << endl;
    }
  }
  return 0;
}
