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;
auto start = std::chrono::system_clock::now();
inline void skj() {
  std::chrono::time_point<std::chrono::system_clock> end;
  end = std::chrono::system_clock::now();
  std::chrono::duration<double> elapsed_seconds = end - start;
  std::time_t end_time = std::chrono::system_clock::to_time_t(end);
  cerr << "Time taken " << elapsed_seconds.count() * 1000 << "\n";
}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
struct custom_hash {
  static uint64_t splitmix64(uint64_t x) {
    x += 0x9e3779b97f4a7c15;
    x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
    x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
    return x ^ (x >> 31);
  }
  size_t operator()(uint64_t x) const {
    static const uint64_t FIXED_RANDOM =
        chrono::steady_clock::now().time_since_epoch().count();
    return splitmix64(x + FIXED_RANDOM);
  }
};
inline long long binexp(long long a, long long b, long long m) {
  if (a == 0) {
    return 0;
  }
  long long res = 1;
  a %= m;
  while (b) {
    if (b & 1) res = (res * 1ll * a) % m;
    a = (a * 1ll * a) % m;
    b >>= 1;
  }
  return res;
}
inline long long binmul(long long a, long long b, long long m) {
  a %= m;
  long long res = 0;
  while (b) {
    if (b & 1) {
      res = (res + a) % m;
    }
    a = (a + a) % m;
    b >>= 1;
  }
  return res;
}
inline void solve_me_senpai() {
  long long n;
  cin >> n;
  vector<long long> a(n), b(n);
  map<long long, set<long long> > M;
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
    M[a[i]].insert(i);
  }
  for (long long i = 0; i < n; i++) {
    cin >> b[i];
  }
  for (long long i = n - 1; i >= 0; i--) {
    if ((M[a[i]].find(i) != M[a[i]].end())) M[a[i]].erase(i);
    if (a[i] == b[i]) continue;
    if (!(long long)M[(b[i] - a[i]) / abs(a[i] - b[i])].size()) {
      cout << "NO\n";
      return;
    }
  }
  cout << "YES\n";
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  cin >> t;
  long long a = 1;
  while (t--) {
    solve_me_senpai();
    a++;
  }
  return 0;
}
