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;
using ll = long long;
const long long int INF = INT_MAX, NINF = INT_MIN;
const long long int mod = 1e9 + 7;
void __print(int x) { cerr << x; }
void __print(long x) { cerr << x; }
void __print(long long x) { cerr << x; }
void __print(unsigned x) { cerr << x; }
void __print(unsigned long x) { cerr << x; }
void __print(unsigned long long x) { cerr << x; }
void __print(float x) { cerr << x; }
void __print(double x) { cerr << x; }
void __print(long double x) { cerr << x; }
void __print(char x) { cerr << '\'' << x << '\''; }
void __print(const char* x) { cerr << '\"' << x << '\"'; }
void __print(const string& x) { cerr << '\"' << x << '\"'; }
void __print(bool x) { cerr << (x ? "true" : "false"); }
template <typename T, typename V>
void __print(const pair<T, V>& x) {
  cerr << '{';
  __print(x.first);
  cerr << ',';
  __print(x.second);
  cerr << '}';
}
template <typename T>
void __print(const T& x) {
  int f = 0;
  cerr << '{';
  for (auto& i : x) cerr << (f++ ? "," : ""), __print(i);
  cerr << "}";
}
void _print() { cerr << "]\n"; }
template <typename T, typename... V>
void _print(T t, V... v) {
  __print(t);
  if (sizeof...(v)) cerr << ", ";
  _print(v...);
}
template <typename T>
void print(vector<T> v) {
  for (T i : v) cout << i << " ";
  cout << '\n';
}
template <typename T>
void print(vector<vector<T>>& v) {
  for (vector<T>& vv : v) {
    for (T& i : vv) cout << i << " ";
    cout << '\n';
  }
}
template <typename T>
void read(vector<T>& v) {
  for (T& i : v) cin >> i;
}
template <typename T>
void read(T&& t) {
  cin >> t;
}
template <typename T, typename... Args>
void read(T&& t, Args&&... args) {
  cin >> t;
  read(forward<Args>(args)...);
}
template <typename T>
void print(T&& t) {
  cout << t << '\n';
}
template <typename T, typename... Args>
void print(T&& t, Args&&... args) {
  cout << t << " ";
  print(forward<Args>(args)...);
}
bool checkPrime(ll n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (ll i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
ll binpow(ll a, ll b, ll mod) {
  a %= mod;
  ll res = 1;
  while (b > 0) {
    if (b & 1) res = (res * a) % mod;
    a = (a * a) % mod;
    b >>= 1;
  }
  return res;
}
ll to_int(string str) {
  ll n = str.size(), num = 0;
  for (int i = 0; i < n; i++) {
    num = num * 10 + (str[i] - '0');
  }
  return num;
}
ll gcd(ll a, ll b) {
  if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
ll lcm(ll a, ll b) { return (a / gcd(a, b) * b); }
ll sumofDigits(ll n) {
  ll sum = 0;
  while (n > 0) {
    sum += n % 10;
    n /= 10;
  }
  return sum;
}
bool isPalindrome(string str) {
  int n = str.size();
  for (int i = 0; i < n; i++) {
    if (str[i] != str[n - i - 1]) {
      return false;
    }
  }
  return true;
}
string toBinary(ll n) {
  string str = "";
  while (n > 0) {
    str += n % 2 + '0';
    n /= 2;
  }
  reverse(str.begin(), str.end());
  return str;
}
ll toDecimal(string str) {
  ll n = str.size(), ans = 0;
  for (ll i = n - 1; i >= 0; i--) {
    if (str[i] == '1') {
      ans += pow(2, n - i - 1);
    }
  }
  return ans;
}
ll modInverse(ll n, ll mod) { return binpow(n, mod - 2, mod); }
ll C(ll n, ll r, ll p) {
  if (r == 0 or n == r) return 1;
  ll fac[n + 1];
  fac[0] = 1;
  for (ll i = 1; i <= n; i++) {
    fac[i] = fac[i - 1] * i;
    fac[i] %= p;
  }
  return (fac[n] * modInverse(fac[r], p) % p * modInverse(fac[n - r], p) % p) %
         p;
}
set<int> storePrimes;
void seive() {
  map<int, int> is_prime;
  for (int i = 2; i * i <= 1000000; i++) {
    if (is_prime[i] == 0)
      for (int j = i * i; j <= 100000; j += i) {
        is_prime[j] = 1;
      }
  }
  for (int i = 2; i <= 1000000; i++) {
    if (is_prime[i] == 0) {
      storePrimes.insert(i);
    }
  }
}
bool isPowerof2(ll n) {
  if ((n & (n - 1)) == 0) return true;
  return false;
}
ll ceil(ll a, ll b) { return (a + b - 1) / b; }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  ;
  int t;
  cin >> t;
  while (t--) {
    ll n;
    cin >> n;
    vector<ll> a(n), b(n);
    for (ll i = 0; i < n; i++) {
      cin >> a[i];
    }
    for (ll i = 0; i < n; i++) {
      cin >> b[i];
    }
    if (a[0] != b[0]) {
      print("NO");
      continue;
    }
    ll pflag = 0, nflag = 0, ok = 1;
    for (ll i = 0; i < n; i++) {
      if (i == 0)
        goto hell;
      else
        goto nice;
    nice:
      if (a[i] != b[i]) {
        if (b[i] > a[i]) {
          if (pflag == 0) {
            ok = 0;
          }
        }
        if (b[i] < a[i]) {
          if (nflag == 0) {
            ok = 0;
          }
        }
      }
    hell:
      if (a[i] > 0) pflag = 1;
      if (a[i] < 0) nflag = 1;
    }
    if (ok) {
      print("YES");
    } else {
      print("NO");
    }
  }
}
