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;
template <typename T>
T getint() {
  T x = 0, p = 1;
  char ch;
  do {
    ch = getchar();
  } while (ch <= ' ');
  if (ch == '-') p = -1, ch = getchar();
  while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
  return x * p;
}
template <typename T1, typename T2>
bool umin(T1& x, const T2& y) {
  if (x > y) return x = y, true;
  return false;
}
template <typename T1, typename T2>
bool umax(T1& x, const T2& y) {
  if (x < y) return x = y, true;
  return false;
}
const int maxn = (int)3e5 + 10;
const int inf = (int)1e9 + 5;
const int mod = 998244353;
const long long llinf = (long long)1e18 + 5;
const double pi = acos(-1.0);
int main() {
  ios_base::sync_with_stdio(false);
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    vector<int> a(n), b(n);
    for (int& x : a) cin >> x;
    for (int& x : b) cin >> x;
    bool any_plus = false;
    bool any_minus = false;
    bool any_positive = false;
    bool any_negative = false;
    bool can = true;
    for (int i = 0; i < n; ++i) {
      const auto Can = [&](int a, int b) {
        const bool can_plus = any_plus | (any_positive & any_minus);
        const bool can_minus = any_minus | (any_negative & any_plus);
        int diff = b - a;
        if (diff == 0) return true;
        if (diff < 0) return can_minus;
        return can_plus;
      };
      if (!Can(a[i], b[i])) {
        can = false;
        break;
      }
      if (b[i] == 1) any_plus = true;
      if (b[i] == -1) any_minus = true;
      if (b[i] > 0) any_positive = true;
      if (b[i] < 0) any_negative = true;
      if (a[i] == 1) any_plus = true;
      if (a[i] == -1) any_minus = true;
      if (a[i] > 0) any_positive = true;
      if (a[i] < 0) any_negative = true;
    }
    cout << (can ? "YES" : "NO") << '\n';
  }
  return 0;
}
