Problem: On a strip of land of length n there are k air conditioners: the i-th air conditioner is placed in cell a_i (1 ≤ a_i ≤ n). Two or more air conditioners cannot be placed in the same cell (i.e. all a_i are distinct).

Each air conditioner is characterized by one parameter: temperature. The i-th air conditioner is set to the temperature t_i.

<image> Example of strip of length n=6, where k=2, a=[2,5] and t=[14,16].

For each cell i (1 ≤ i ≤ n) find it's temperature, that can be calculated by the formula $$$min_{1 ≤ j ≤ k}(t_j + |a_j - i|),$$$

where |a_j - i| denotes absolute value of the difference a_j - i.

In other words, the temperature in cell i is equal to the minimum among the temperatures of air conditioners, increased by the distance from it to the cell i.

Let's look at an example. Consider that n=6, k=2, the first air conditioner is placed in cell a_1=2 and is set to the temperature t_1=14 and the second air conditioner is placed in cell a_2=5 and is set to the temperature t_2=16. In that case temperatures in cells are:

  1. temperature in cell 1 is: min(14 + |2 - 1|, 16 + |5 - 1|)=min(14 + 1, 16 + 4)=min(15, 20)=15; 
  2. temperature in cell 2 is: min(14 + |2 - 2|, 16 + |5 - 2|)=min(14 + 0, 16 + 3)=min(14, 19)=14; 
  3. temperature in cell 3 is: min(14 + |2 - 3|, 16 + |5 - 3|)=min(14 + 1, 16 + 2)=min(15, 18)=15; 
  4. temperature in cell 4 is: min(14 + |2 - 4|, 16 + |5 - 4|)=min(14 + 2, 16 + 1)=min(16, 17)=16; 
  5. temperature in cell 5 is: min(14 + |2 - 5|, 16 + |5 - 5|)=min(14 + 3, 16 + 0)=min(17, 16)=16; 
  6. temperature in cell 6 is: min(14 + |2 - 6|, 16 + |5 - 6|)=min(14 + 4, 16 + 1)=min(18, 17)=17. 



For each cell from 1 to n find the temperature in it.

Input

The first line contains one integer q (1 ≤ q ≤ 10^4) — the number of test cases in the input. Then test cases follow. Before each test case, there is an empty line.

Each test case contains three lines. The first line contains two integers n (1 ≤ n ≤ 3 ⋅ 10^5) and k (1 ≤ k ≤ n) — the length of the strip of land and the number of air conditioners respectively.

The second line contains k integers a_1, a_2, …, a_k (1 ≤ a_i ≤ n) — positions of air conditioners on the strip of land.

The third line contains k integers t_1, t_2, …, t_k (1 ≤ t_i ≤ 10^9) — temperatures of air conditioners.

It is guaranteed that the sum of n over all test cases does not exceed 3 ⋅ 10^5.

Output

For each test case output n integers separated by space: temperatures of air in cells.

Example

Input


5

6 2
2 5
14 16

10 1
7
30

5 5
3 1 4 2 5
3 1 4 2 5

7 1
1
1000000000

6 3
6 1 3
5 5 5


Output


15 14 15 16 16 17 
36 35 34 33 32 31 30 31 32 33 
1 2 3 4 5 
1000000000 1000000001 1000000002 1000000003 1000000004 1000000005 1000000006 
5 6 5 6 6 5 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std ;

#define int long long
#define ull unsigned long long
#define N 100001
#define endl "\n"


/*
vector <bool> prime(N + 1, true) ;
vector <ull> fac(N) ;
vector <ull> inverse_fac(N) ;


bool check_prime(int n){
	if(n == 1)
		return false ;
	if(n % 2 == 0)
		return false ;
	for(int i = 3; i <= sqrt(n); i += 2){
		if(n % i == 0)
			return false ;
	}
	return true ;
}

void SieveOfEratosthenes(int n) {
	prime[0] = false ;
	prime[1] = false ;
	for (int i = 2; i * i <= n; i++) {
		if (prime[i] == true) {
			for (int j = i * i; j <= n; j += i)
				prime[j] = false;
		}
	}
	return ;
}

void calc_fact(ull mod) {
	fac[0] = 1;
	for (int i = 1; i < N; i++)
		fac[i] = (fac[i - 1] * i) % mod;
}

ull mod_expo(ull a, ull n, ull mod) {
	ull res{1} ;
	a = a % mod;
	while (n) {
		if (n & 1)
			res = (res * a) % mod;
		n = n >> 1;
		a = (a * a) % mod ;
	}
	return res % mod;
}

ull modInverse(ull n, ull mod) {
	return mod_expo(n, mod - 2, mod);
}

void calc_inverse_fact(ull mod) {
	inverse_fac[N - 1] = modInverse(fac[N - 1], mod) ;
	for (int i = N - 2; i >= 0; i--) {
		inverse_fac[i] = (inverse_fac[i + 1] * (i + 1)) % mod ;
	}
}


ull nCr_mod_fermat1(ull n, ull r, ull mod) {
	if (n < r)
		return 0;
	if (r == 0)
		return 1;
	ull res{fac[n]} ;
	res = (res * modInverse(fac[r], mod)) % mod ;
	res = (res * modInverse(fac[n - r], mod)) % mod ;
	return res ;
}

ull nCr_mod_fermat2(ull n, ull r, ull mod) {
	if (n < r)
		return 0;
	if (r == 0)
		return 1;
	ull res{fac[n]} ;
	res = (res * inverse_fac[r]) % mod ;
	res = (res * inverse_fac[n - r]) % mod ;
	return res ;
}


int extended_gcd(int a, int b, int *x, int *y) {
	if (a == 0) {
		*x = 0;
		*y = 1;
		return b;
	}
	int x1{}, y1{} ;
	int gcd = extended_gcd(b % a, a, &x1, &y1);
	*x = y1 - (b / a) * x1;
	*y = x1;
	return gcd;
}


vector <int> find_next_greater_element(vector <int> &vec) {
	int n{} ;
	n = vec.size() ;
	vector <int> res(n, -1) ;
	stack <int> st{} ;
	st.push(n - 1);
	for (int i = n - 2; i >= 0; i--) {
		while (!st.empty() && vec[st.top()] <= vec[i])
			st.pop();
		if (!st.empty())
			res[i] = st.top() ;
		st.push(i);
	}
	return res ;
}

vector <int> find_prev_smaller_element(vector <int> &vec) {
	int n{} ;
	n = vec.size() ;
	vector <int> res(n, -1) ;
	stack <int> st{} ;
	st.push(0);
	for (int i = 1; i < n; i++) {
		while (!st.empty() && vec[st.top()] >= vec[i])
			st.pop();
		if (!st.empty())
			res[i] = st.top() ;
		st.push(i);
	}
	return res ;
}

struct custom_hash {
	static uint64_t splitmix64(uint64_t x) {
		x += 0x9e3779b97f4a7c15;
		x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
		x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
		return x ^ (x >> 31);
	}

	size_t operator()(uint64_t x) const {
		static const uint64_t FIXED_RANDOM = chrono::steady_clock::now().time_since_epoch().count();
		return splitmix64(x + FIXED_RANDOM);
	}
};

// user defined comp for sets etc.
struct comp {
	bool operator()(const pair <int, int> &p1, const pair <int, int> &p2) {
		if (p1.first != p2.first)
			return p1.first < p2.first ;
		return p1.second > p2.second ;
	}
} ;


#define INF 1e18


// Counting no of inversions using a segment tree
vector <int> st(400004) ;              // sum of a segment

void build_st(int si, int ss, int se, vector <int> &vis) {
	if (ss == se) {
		st[si] = vis[ss] ;
		return ;
	}
	int mid{} ;
	mid = (ss + (se - ss) / 2) ;
	build_st(2 * si, ss, mid, vis) ;
	build_st(2 * si + 1, mid + 1, se, vis) ;
	st[si] = st[2 * si] + st[2 * si + 1] ;
}

int query_st(int si, int ss, int se, int qs, int qe, vector <int> &vis) {
	if (qs > se || qe < ss)
		return 0 ;
	else if (qs <= ss && qe >= se)
		return st[si] ;
	int mid{} ;
	mid = ss + (se - ss) / 2 ;
	int l{}, r{} ;
	l = query_st(2 * si, ss, mid, qs, qe, vis) ;
	r = query_st(2 * si + 1, mid + 1, se, qs, qe, vis) ;
	return l + r ;
}

void point_update_st(int si, int ss, int se, int qi, vector <int> &vis) {
	if (ss == se) {
		st[si] = vis[se] ;
		return ;
	}
	int mid{} ;
	mid = ss + (se - ss) / 2 ;
	if (qi <= mid)
		point_update_st(2 * si, ss, mid, qi, vis) ;
	else
		point_update_st(2 * si + 1, mid + 1, se, qi, vis) ;
	st[si] = st[2 * si] + st[2 * si + 1] ;
}
*/



int32_t main() {
#ifndef ONLINE_JUDGE
	freopen("input1.txt", "r", stdin) ;

	freopen("output1.txt", "w", stdout) ;
#endif
	ios_base::sync_with_stdio(false) ;
	cin.tie(NULL) ;
	int t{} ;
	cin >> t ;
	while (t--) {
		int n{}, k{} ;
		cin >> n >> k ;
		vector <int> vec1(k, 0), vec2(k, 0) ;
		for (int i = 0; i < k; i++)
			cin >> vec1[i] ;
		for (int i = 0; i < k; i++)
			cin >> vec2[i] ;
		vector <int> v(n, -1) ;
		for (int i = 0; i < k; i++) {
			v[vec1[i] - 1] = vec2[i] ;
		}
		vector <int> ans(n, LLONG_MAX / 2) ;
		vector <int> left(n, LLONG_MAX / 2), right(n, LLONG_MAX / 2) ;
		for (int i = 0; i < n; i++) {
			if (v[i] != -1)
				ans[i] = v[i] ;
		}
		if (v[0] != -1)
			left[0] = v[0] ;
		for (int i = 1; i < n; i++) {
			left[i] = min(left[i], left[i - 1] + 1) ;
			if (v[i] != -1)
				left[i] = min(left[i], v[i]) ;
		}
		if (v[n - 1] != -1)
			right[n - 1] = v[n - 1] ;
		for (int i = n - 2; i >= 0; i--) {
			right[i] = min(right[i], right[i + 1] + 1) ;
			if (v[i] != -1)
				right[i] = min(right[i], v[i]) ;
		}
		for (int i = 0; i < n; i++) {
			ans[i] = min({ans[i], left[i], right[i]}) ;
		}
		for (auto elem : ans)
			cout << elem << " " ;
		cout << endl ;
	}
	return 0 ;
}