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;
static char inp[1024], *sde = inp + 1024;
static const char *sp = sde;
struct __X {
  bool d[256], dd[256], kk = 1, s = 1;
  __X() {
    for (int i = 48; i < 58; d[i++] = 1)
      ;
    d[45] = dd[32] = dd[10] = dd[13] = dd[9] = 1;
  }
  __attribute__((always_inline)) inline void __ra() {
    size_t r = sde - sp;
    if (r < 24) {
      memcpy(inp, sp, r);
      size_t sz = fread(inp + r, 1, 1024 - r, stdin);
      sp = inp;
      sde = inp + r + sz;
      if (sde != inp + 1024) *sde = 0;
    }
    if (sp == sde) kk = 0;
  }
  __attribute__((always_inline)) inline char P() {
    if (!(sde - sp)) __ra();
    return *(sp++);
  }
  __attribute__((always_inline)) inline __X &operator,(char &x) {
    if (kk)
      for (x = P(); kk && dd[x]; x = P())
        ;
    return *this;
  }
  __attribute__((always_inline)) inline __X &operator,(char *x) {
    if (kk) {
      char _t = P();
      for (; kk && dd[_t]; _t = P())
        ;
      if (_t)
        for (; _t && !dd[_t]; _t = P()) *x++ = _t;
      *x = 0;
    }
    return *this;
  }
  __attribute__((always_inline)) inline __X &operator,(string &x) {
    x = "";
    if (kk) {
      char _t = P();
      for (; kk && dd[_t]; _t = P())
        ;
      if (_t)
        for (; _t && !dd[_t]; _t = P()) x += _t;
    }
    return *this;
  }
  template <class T>
  __attribute__((always_inline)) inline __X &operator,(T &x) {
    if (kk) {
      x = s = 0;
      __ra();
      while (!d[*sp]) __ra(), ++sp;
      if (*sp == '-')
        s = 1;
      else
        x = (*sp & 15);
      ++sp;
      while (d[*sp]) x = (x << 1) + (x << 3) + (*(sp++) & 15);
      if (s) x = -x;
    }
    return *this;
  }
} in;
int main() {
  int t, a[100000], b[100000], n, i, fp, fn;
  bool st;
  in, t;
  while (t--) {
    in, n;
    st = false;
    fp = fn = n;
    for (i = 0; i < n; ++i) in, a[i];
    for (i = 0; i < n; ++i) in, b[i];
    for (i = 0; i < n; ++i)
      if (a[i] > 0) {
        fp = i;
        break;
      }
    for (i = 0; i < n; ++i)
      if (a[i] < 0) {
        fn = i;
        break;
      }
    for (i = 0; i < n; ++i) {
      if (a[i] < b[i]) {
        if (fp >= i) {
          st = true;
          break;
        }
      } else if (a[i] > b[i]) {
        if (fn >= i) {
          st = true;
          break;
        }
      }
    }
    cout << (st ? "NO\n" : "YES\n");
  }
}
