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;
const double PI = 2 * acos(0.0);
const int Mod = 1E9 + 7;
inline void yes() { std::cout << "Yes\n"; }
inline void no() { std::cout << "No\n"; }
using namespace std;
long long int isPrime(long long int num) {
  long long int flag = 0;
  for (long long int i = 1; i <= num; i++) {
    if (num % i == 0) {
      flag++;
      if (flag > 3) break;
    }
  }
  if (flag == 3)
    return 1;
  else
    return 0;
}
long long int binomialCoeff(long long int n, long long int k) {
  long long int C[k + 1];
  memset(C, 0, sizeof(C));
  C[0] = 1;
  for (long long int i = 1; i <= n; i++) {
    for (long long int j = min(i, k); j > 0; j--) C[j] = C[j] + C[j - 1];
  }
  return C[k];
}
long long int factorial(long long int n) {
  if (n == 0) return 1;
  return n * factorial(n - 1);
}
bool isPalindrome(string str) {
  int l = 0;
  int h = str.length() - 1;
  while (h > l) {
    if (str[l++] != str[h--]) {
      return false;
    }
  }
  return true;
}
bool check(long long int n) {
  double y = sqrt(n);
  if (y - floor(y) == 0)
    return 1;
  else
    return 0;
}
bool isprime(long long int N) {
  if (N < 2 || (!(N & 1) && N != 2)) return 0;
  for (int i = 3; i * i <= N; i += 2) {
    if (!(N % i)) return 0;
  }
  return 1;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int x, y, cnt = 0, l, o, t, n, sum = 0, lasum = 0, ans = 0, las = 0,
                      flag = 0, ma = LLONG_MAX, av1, av2, sum1 = 0, sum2 = 0;
  cin >> t;
  while (t--) {
    cin >> n;
    long long int a[n], b[n], c[n];
    for (long long int i = (0); i < (n); i++) {
      cin >> a[i];
    }
    for (long long int i = (0); i < (n); i++) {
      cin >> b[i];
      c[i] = a[i] - b[i];
    }
    long long int pos = 0, neg = 0;
    flag = 0;
    for (long long int i = (0); i < (n); i++) {
      if (c[i] != 0) {
        if (c[i] < 0) {
          if (pos == 0) {
            cout << "NO" << '\n';
            flag = 1;
            break;
          }
        } else {
          if (neg == 0) {
            cout << "NO" << '\n';
            flag = 1;
            break;
          }
        }
      }
      if (a[i] > 0) pos = 1;
      if (a[i] < 0) neg = 1;
    }
    if (flag == 1) continue;
    cout << "YES" << '\n';
  }
}
