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 pii = pair<int, int>;
const int MOD = 1000000007;
const int INF = 2000000000;
const long long DESPACITO = 1000000000000000000;
namespace minmax {
template <class T>
T max(T&& A) {
  return A;
}
template <class T>
T min(T&& A) {
  return A;
}
template <class T>
T max(T&& A, T&& B) {
  return A > B ? A : B;
}
template <class T>
T chmin2(T&& A, T&& B) {
  return A < B ? A : B;
}
template <class T, class... Ts>
T max(T&& A, Ts&&... ts) {
  T B = max(ts...);
  return A > B ? A : B;
}
template <class T, class... Ts>
T min(T&& A, Ts&&... ts) {
  T B = min(ts...);
  return A < B ? A : B;
}
template <class T, class... Ts>
T chmax(T&& A, Ts&&... ts) {
  A = max(A, ts...);
  return A;
}
template <class T, class... Ts>
T chmin(T&& A, Ts&&... ts) {
  A = min(A, ts...);
  return A;
}
template <class T1, class T2>
void chmin2(T1&& A, T2&& Aa, T1&& B, T2&& Bb) {
  if (B < A) A = B, Aa = Bb;
}
template <class T1, class T2>
void chmax2(T1&& A, T2&& Aa, T1&& B, T2&& Bb) {
  if (B > A) A = B, Aa = Bb;
}
}  // namespace minmax
namespace input {
template <class T>
void re(T&& Xx) {
  cin >> Xx;
}
template <class T1, class T2>
void re(pair<T1, T2>& Pp) {
  re(Pp.first);
  re(Pp.second);
}
template <class T>
void re(vector<T>& Aa) {
  for (int i = 0; i < (int)(Aa).size(); i++) re(Aa[i]);
}
template <class T1, class T2>
void rea(T1&& Aa, T2 t) {
  for (int i = 0; i < t; i++) re(Aa[i]);
}
template <class T1, class T2, class... Ts>
void rea(T1&& Aa, T2 t, Ts&&... ts) {
  rea(Aa, t);
  rea(ts...);
}
template <class T1, class T2>
void rea1(T1&& Aa, T2 t) {
  for (int i = 1; i <= t; i++) re(Aa[i]);
}
template <class T1, class T2, class... Ts>
void rea1(T1&& Aa, T2 t, Ts&... ts) {
  rea1(Aa, t);
  rea1(ts...);
}
template <class T, class... Ts>
void re(T&& t, Ts&... ts) {
  re(t);
  re(ts...);
}
}  // namespace input
namespace output {
void pr(int32_t Xx) { cout << Xx; }
void pr(bool Xx) { cout << Xx; }
void pr(long long Xx) { cout << Xx; }
void pr(long long unsigned Xx) { cout << Xx; }
void pr(double Xx) { cout << Xx; }
void pr(char Xx) { cout << Xx; }
void pr(const string& Xx) { cout << Xx; }
void pr(const char* Xx) { cout << Xx; }
void pr(const char* Xx, size_t len) { cout << string(Xx, len); }
void ps() { cout << "\n"; }
void pn() {}
void pw() { pr(" "); }
void pc() {
  pr("]");
  ps();
}
template <class T1, class T2>
void pr(const pair<T1, T2>& Xx) {
  pr(Xx.first);
  pw();
  pr(Xx.second);
}
template <class T>
void pr(const T&);
bool parse(const char* t) {
  if (t == "\n") return true;
  return false;
}
template <class T>
bool parse(T&& t) {
  return false;
}
template <class T1, class T2>
bool parsepair(const pair<T1, T2>& Xx) {
  return true;
}
template <class T>
bool parsepair(T&& t) {
  return false;
}
template <class T1, class T2>
void psa(T1&& Aa, T2 t) {
  for (int i = 0; i < t; i++) pr(Aa[i]), pw();
  ps();
}
template <class T1, class T2>
void pna(T1&& Aa, T2 t) {
  for (int i = 0; i < t; i++) pr(Aa[i]), ps();
}
template <class T1, class T2>
void psa2(T1&& Aa, T2 t1, T2 t2) {
  for (int i = 0; i < t1; i++) {
    for (int j = 0; j < t2; j++) pr(Aa[i][j]), pw();
    ps();
  }
}
template <class T>
void pr(const T& Xx) {
  if (!(int)(Xx).size()) return;
  bool fst = 1;
  bool op = 0;
  if (parsepair(*Xx.begin())) op = 1;
  for (const auto& Aa : Xx) {
    if (!fst) pw();
    if (op) pr("{");
    pr(Aa), fst = 0;
    if (op) pr("}");
  }
}
template <class T, class... Ts>
void pr(const T& t, const Ts&... ts) {
  pr(t);
  pr(ts...);
}
template <class T, class... Ts>
void ps(const T& t, const Ts&... ts) {
  pr(t);
  if (sizeof...(ts)) {
    if (!parse(t)) pw();
  }
  ps(ts...);
}
template <class T>
void pn(const T& t) {
  for (const auto& Aa : t) ps(Aa);
}
template <class T, class... Ts>
void pw(const T& t, const Ts&... ts) {
  pr(t);
  if (sizeof...(ts)) pw();
  pw(ts...);
}
template <class T, class... Ts>
void pc(const T& t, const Ts&... ts) {
  bool op = 0;
  if (parsepair(t)) op = 1;
  if (op) pr("{");
  pr(t);
  if (op) pr("}");
  if (sizeof...(ts)) pr(", ");
  pc(ts...);
}
namespace trace {
template <class T1, class T2>
void __f(const char* name, const T1& Xx, const T2& y) {
  pr("[", y, "] = ");
  ps(Xx);
}
template <class T1, class T2, class... Ts>
void __f(const char* name, const T1& Xx, const T2& y, const Ts&... rest) {
  const char* open = strchr(name, '[');
  pr("[", y, "]");
  __f(open + 1, Xx, rest...);
}
template <class T1, class T2, class... Ts>
void __f0(const char* name, const T1& Xx, const T2& y, const Ts&... rest) {
  const char* open = strchr(name, '[');
  pr(name, size_t(open - name));
  __f(name, Xx, y, rest...);
}
template <class T>
void __fn(int n, const char* name, const T& Xx) {
  for (int i = 0; i < n; i++) pr(name), __f(name, Xx[i], i);
}
}  // namespace trace
}  // namespace output
using namespace minmax;
using namespace input;
using namespace output;
using namespace output::trace;
const int N = 2e5 + 5;
int a[N], b[N];
int solve() {
  int i, n;
  re(n);
  rea(a, n, b, n);
  int cnt[3] = {};
  for (i = 0; i < n; i++) cnt[a[i] + 1]++;
  for (i = n - 1; ~i; i--) {
    cnt[a[i] + 1]--;
    if (a[i] == b[i]) continue;
    if (a[i] < b[i] and !cnt[1 + 1]) return ps("NO"), 0;
    if (a[i] > b[i] and !cnt[-1 + 1]) return ps("NO"), 0;
  }
  ps("YES");
  return 0;
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int Q;
  for (cin >> Q; Q; Q--) solve();
  return 0;
}
