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 int MOD = 1e9 + 7;
const int nax = 1e6 + 10;
vector<int> tree(4 * nax);
int i, j, x, y, k;
bool sortbysec(const pair<long long, long long> &a,
               const pair<long long, long long> &b) {
  return (a.second < b.second);
}
class bitmask {
 public:
  void PowerSet(char *set, int set_size) {
    unsigned int pow_set_size = pow(2, set_size);
    int counter, j;
    for (counter = 0; counter < pow_set_size; counter++) {
      for (j = 0; j < set_size; j++) {
        if (counter & (1 << j)) cout << set[j];
      }
      cout << endl;
    }
  }
};
long long powerLL(long long x, long long n) {
  long long result = 1;
  while (n) {
    if (n & 1) result = result * x % MOD;
    n = n / 2;
    x = x * x % MOD;
  }
  return result;
}
bool sortbylength(string &s1, string &s2) { return s1.size() >= s2.size(); }
void showstack(stack<char> s) {
  while (!s.empty()) {
    cout << '\t' << s.top();
    s.pop();
  }
  cout << '\n';
}
bool correctsequense(string s) {
  long long prefix = 0;
  long long i = 0;
  for (i = 0; i < s.length(); i++) {
    if (s[i] == '(')
      prefix++;
    else
      prefix--;
    if (prefix < 0) return false;
  }
  return true;
}
bool ispalindrome(string s) {
  string temp = s;
  reverse(temp.begin(), temp.end());
  if (temp == s)
    return true;
  else
    return false;
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n, t, m, y;
  cin >> t;
  while (t--) {
    cin >> n;
    long long i = 0, flag = 0;
    vector<long long> a, b;
    vector<int> v(3, 0);
    for (i = 0; i < n; i++) {
      cin >> y;
      a.push_back(y);
    }
    for (i = 0; i < n; i++) {
      cin >> y;
      b.push_back(y);
    }
    for (i = 0; i < n; i++) {
      if (a[i] != b[i]) {
        if (b[i] > 0 && v[1] != 1 || (b[i] < 0 && v[2] != 1)) {
          flag = 1;
          break;
        } else if (b[i] == 0) {
          if (a[i] > 0 && v[2] != 1) {
            flag = 1;
            break;
          } else if (a[i] < 0 && v[1] != 1) {
            flag = 1;
            break;
          }
        }
      }
      if (a[i] == 1)
        v[1] = 1;
      else if (a[i] == -1)
        v[2] = 1;
      else
        v[0] = 1;
    }
    if (flag == 0)
      cout << "yes" << endl;
    else
      cout << "no" << endl;
  }
}
