Kind Anton - VietMX's Blog

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:

Choose any pair of indexes $(i, j)$ such that $1 \le i < j \le n$ . It is possible to choose the same pair $(i, j)$ more than once.
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 $t$ ( $1 \le t \le 10000$ ). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ( $1 \le n \le 10^5$ ) — the length of arrays.

The second line of each test case contains $n$ integers $a_1, a_2, \dots, a_n$ ( $-1 \le a_i \le 1$ ) — elements of array $a$ . There can be duplicates among elements.

The third line of each test case contains $n$ integers $b_1, b_2, \dots, b_n$ ( $-10^9 \le b_i \le 10^9$ ) — 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 $[1, -1, 0] \to [1, -1, -2] \to [1, 1, -2]$

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 $b$ .

Solution:

#include <bits/stdc++.h>
using namespace std;
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=n-1;i>=a;i--)
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
typedef vector<int> VI;
typedef long long ll;
typedef pair<int,int> PII;
typedef double db;
mt19937 mrand(random_device{}());
const ll mod=1000000007;
int rnd(int x) { return mrand() % x;}
ll powmod(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
ll gcd(ll a,ll b) { return b?gcd(b,a%b):a;}
// head
 
const int N=101000;
int n,a[N],b[N],_;
 
bool solve() {
scanf("%d",&n);
rep(i,0,n) {
scanf("%d",a+i);
}
bool p1=0,n1=0;
rep(i,0,n) {
scanf("%d",b+i);
}
rep(i,0,n) {
int x=b[i]-a[i];
if (p1==0&&n1==0) {
if (x!=0) return 0;
}
if (p1==0) {
if (x>0) return 0;
}
if (n1==0) {
if (x<0) return 0;
        }
        if (a[i]==1) p1=1;
        if (a[i]==-1) n1=1;
    }
    return 1;
}
 
int main() {
    for (scanf("%d",&_);_;_--) {
        puts(solve()?"YES":"NO");
    }
}