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;
__attribute__((constructor)) static void fast_cin() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
}
using ll = int64_t;
using ull = uint64_t;
using mp = pair<ll, ll>;
using mt = tuple<ll, ll, ll>;
template <typename T, typename R>
istream &operator>>(istream &fin, pair<T, R> &p) {
  fin >> p.first >> p.second;
  return fin;
}
template <typename T, typename R>
ostream &operator<<(ostream &fout, const pair<T, R> &p) {
  fout << "{" << p.first << ", " << p.second << "} ";
  return fout;
}
template <typename T>
ostream &operator<<(ostream &fout, const vector<T> &vec) {
  for (size_t i = 0; i < vec.size(); i++) fout << vec[i] << ' ';
  return fout;
}
template <typename T>
istream &operator>>(istream &fin, vector<T> &vec) {
  for (size_t i = 0; i < vec.size(); i++) fin >> vec[i];
  return fin;
}
vector<string> split(const string &s, char c) {
  vector<string> v;
  ull st = 0, fn = 0, lvl = 0;
  for (; fn < s.size(); fn++) {
    if (s[fn] == c && lvl == 0) {
      v.push_back(s.substr(st, fn - st));
      st = fn + 1;
    }
    if (s[fn] == '(') {
      lvl++;
    }
    if (s[fn] == ')') {
      lvl--;
    }
  }
  v.push_back(s.substr(st, fn - st));
  return v;
}
void pr_var(vector<string>::iterator it) {
  (void)it;
  cerr << endl;
}
template <typename T, typename... Args>
void pr_var(vector<string>::iterator it, T a, Args... args) {
  cerr << *it << "=" << a << "  ";
  pr_var(++it, args...);
}
ll div_ceil(ll n, ll d) {
  ll res = n / d;
  if (n >= 0) res += n % d ? 1 : 0;
  return res;
}
ll div_floor(ll n, ll d) {
  ll res = n / d;
  if (n < 0) res -= n % d ? 1 : 0;
  return res;
}
ll gcd(ll a, ll b) {
  a = llabs(a);
  b = llabs(b);
  while (a && b) a > b ? a %= b : b %= a;
  return a + b;
}
ll lcm(ll a, ll b) {
  ll c = gcd(a, b);
  return c == 0 ? 0 : llabs(a / c * b);
}
const double PI = acos(-1.0);
constexpr double EPS = 1e-9;
constexpr ll INF = numeric_limits<ll>::max();
constexpr ll MOD = 1e9 + 7;
int main() {
  ll t;
  cin >> t;
  while (t--) {
    ll n;
    cin >> n;
    vector<ll> a(n);
    cin >> a;
    vector<ll> b(n);
    cin >> b;
    ll plus = 0, minus = 0;
    for (ll i = 0; i < n; i++) {
      if (a[i] == 1)
        plus++;
      else if (a[i] == -1)
        minus++;
    }
    do {
    } while (0);
    int no = 0;
    for (ll i = n - 1; i >= 0; i--) {
      if (a[i] == 1)
        plus--;
      else if (a[i] == -1)
        minus--;
      ll need = b[i] - a[i];
      if (need == 0)
        continue;
      else if (need > 0) {
        if (!plus) no = 1;
      } else if (need < 0) {
        if (!minus) no = 1;
      }
    }
    do {
    } while (0);
    if (no)
      cout << "NO\n";
    else
      cout << "YES\n";
  }
  return 0;
}
