#include <bits/stdc++.h>
using namespace std;
void print(vector<long long int> &arr) {
  for (long long int i = 0; i < arr.size(); i++) cout << arr[i] << " ";
  cout << endl;
}
void print(long long int arr[], long long int n) {
  for (long long int i = 0; i < n; i++) {
    cout << arr[i] << " ";
  }
  cout << endl;
}
long long int max(vector<long long int> &arr) {
  long long int maxi = INT_MIN;
  for (int i = 0; i < arr.size(); i++) {
    if (arr[i] > maxi) maxi = arr[i];
  }
  return maxi;
}
long long int min(vector<long long int> &arr) {
  long long int maxi = INT_MAX;
  for (int i = 0; i < arr.size(); i++) {
    if (arr[i] < maxi) maxi = arr[i];
  }
  return maxi;
}
long long int factorial(long long int x) {
  long long int r = 1;
  while (x) {
    r = r * x;
    x--;
  }
}
long long int gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long int gcdExtended(long long int a, long long int b, long long int *x,
                          long long int *y) {
  if (a == 0) {
    *x = 0, *y = 1;
    return b;
  }
  long long int x1, y1;
  long long int gcd = gcdExtended(b % a, a, &x1, &y1);
  *x = y1 - (b / a) * x1;
  *y = x1;
  return gcd;
}
long long int modInverse(long long int a, long long int m) {
  long long int x, y;
  long long int g = gcdExtended(a, m, &x, &y);
  if (g != 1)
    return INT_MIN;
  else {
    long long int res = (x % m + m) % m;
    return res;
  }
}
int main() {
  long long int t;
  cin >> t;
  while (t--) {
    long long int n;
    cin >> n;
    vector<long long int> a(n);
    vector<long long int> b(n);
    for (int i = 0; i < n; i++) cin >> a[i];
    for (int j = 0; j < n; j++) cin >> b[j];
    bool is1 = false;
    bool is = false;
    bool var = true;
    for (int i = 0; i < n; i++) {
      int diff = b[i] - a[i];
      if (diff > 0 && is1 == true) {
        ;
      } else if (diff < 0 && is == true) {
        ;
      } else if (diff == 0) {
        ;
      } else {
        var = false;
        break;
      }
      if (a[i] == 1) is1 = true;
      if (a[i] == -1) is = true;
    }
    if (var)
      cout << "YES" << endl;
    else
      cout << "NO" << endl;
  }
  return 0;
}

### Prompt

Your challenge is to write a cpp solution to the following problem:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
void print(vector<long long int> &arr) {
  for (long long int i = 0; i < arr.size(); i++) cout << arr[i] << " ";
  cout << endl;
}
void print(long long int arr[], long long int n) {
  for (long long int i = 0; i < n; i++) {
    cout << arr[i] << " ";
  }
  cout << endl;
}
long long int max(vector<long long int> &arr) {
  long long int maxi = INT_MIN;
  for (int i = 0; i < arr.size(); i++) {
    if (arr[i] > maxi) maxi = arr[i];
  }
  return maxi;
}
long long int min(vector<long long int> &arr) {
  long long int maxi = INT_MAX;
  for (int i = 0; i < arr.size(); i++) {
    if (arr[i] < maxi) maxi = arr[i];
  }
  return maxi;
}
long long int factorial(long long int x) {
  long long int r = 1;
  while (x) {
    r = r * x;
    x--;
  }
}
long long int gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long int gcdExtended(long long int a, long long int b, long long int *x,
                          long long int *y) {
  if (a == 0) {
    *x = 0, *y = 1;
    return b;
  }
  long long int x1, y1;
  long long int gcd = gcdExtended(b % a, a, &x1, &y1);
  *x = y1 - (b / a) * x1;
  *y = x1;
  return gcd;
}
long long int modInverse(long long int a, long long int m) {
  long long int x, y;
  long long int g = gcdExtended(a, m, &x, &y);
  if (g != 1)
    return INT_MIN;
  else {
    long long int res = (x % m + m) % m;
    return res;
  }
}
int main() {
  long long int t;
  cin >> t;
  while (t--) {
    long long int n;
    cin >> n;
    vector<long long int> a(n);
    vector<long long int> b(n);
    for (int i = 0; i < n; i++) cin >> a[i];
    for (int j = 0; j < n; j++) cin >> b[j];
    bool is1 = false;
    bool is = false;
    bool var = true;
    for (int i = 0; i < n; i++) {
      int diff = b[i] - a[i];
      if (diff > 0 && is1 == true) {
        ;
      } else if (diff < 0 && is == true) {
        ;
      } else if (diff == 0) {
        ;
      } else {
        var = false;
        break;
      }
      if (a[i] == 1) is1 = true;
      if (a[i] == -1) is = true;
    }
    if (var)
      cout << "YES" << endl;
    else
      cout << "NO" << endl;
  }
  return 0;
}

```