#include <bits/stdc++.h>
const long long INF = 1e18;
const long long MOD = 1000000007;
using namespace std;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int tt;
  cin >> tt;
  while (tt--) {
    int n;
    cin >> n;
    vector<long long> a(n), b(n);
    for (long long &i : a) cin >> i;
    for (long long &i : b) cin >> i;
    if (a[0] != b[0]) {
      cout << "NO\n";
      continue;
    }
    vector<map<int, bool> > maps(n);
    map<int, bool> m;
    m[a[0]] = true;
    maps[0] = m;
    for (int i = 1; i < n; i++) {
      maps[i] = maps[i - 1];
      maps[i][a[i]] = true;
    }
    bool sw = false;
    for (int i = 0; i < n; i++) {
      if (a[i] != b[i]) {
        sw = true;
        break;
      }
    }
    if (!sw) {
      cout << "YES\n";
      continue;
    }
    sw = false;
    for (int i = n - 1; i >= 1; i--) {
      if (a[i] == b[i]) {
        continue;
      }
      if (a[i] > b[i]) {
        if (maps[i - 1][-1] == true) {
          continue;
        } else {
          cout << "NO\n";
          sw = true;
          break;
        }
      }
      if (a[i] < b[i]) {
        if (maps[i - 1][1] == true) {
          continue;
        } else {
          cout << "NO\n";
          sw = true;
          break;
        }
      }
    }
    if (!sw) cout << "YES\n";
  }
}

### Prompt

Please create a solution in cpp 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>
const long long INF = 1e18;
const long long MOD = 1000000007;
using namespace std;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int tt;
  cin >> tt;
  while (tt--) {
    int n;
    cin >> n;
    vector<long long> a(n), b(n);
    for (long long &i : a) cin >> i;
    for (long long &i : b) cin >> i;
    if (a[0] != b[0]) {
      cout << "NO\n";
      continue;
    }
    vector<map<int, bool> > maps(n);
    map<int, bool> m;
    m[a[0]] = true;
    maps[0] = m;
    for (int i = 1; i < n; i++) {
      maps[i] = maps[i - 1];
      maps[i][a[i]] = true;
    }
    bool sw = false;
    for (int i = 0; i < n; i++) {
      if (a[i] != b[i]) {
        sw = true;
        break;
      }
    }
    if (!sw) {
      cout << "YES\n";
      continue;
    }
    sw = false;
    for (int i = n - 1; i >= 1; i--) {
      if (a[i] == b[i]) {
        continue;
      }
      if (a[i] > b[i]) {
        if (maps[i - 1][-1] == true) {
          continue;
        } else {
          cout << "NO\n";
          sw = true;
          break;
        }
      }
      if (a[i] < b[i]) {
        if (maps[i - 1][1] == true) {
          continue;
        } else {
          cout << "NO\n";
          sw = true;
          break;
        }
      }
    }
    if (!sw) cout << "YES\n";
  }
}

```