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 JAVA):

Solution:
import java.util.*;
import java.io.*;

public class E1547 {
	static boolean multipleTC = true;
	FastReader in;
	PrintWriter out;

	static int mod = 1000000007;

	public static void main(String[] args) throws Exception {
		new E1547().run();
	}

	void run() throws Exception {
		in = new FastReader();
		out = new PrintWriter(System.out);
		int T = (multipleTC) ? ni() : 1;
		pre();
		for (int t = 1; t <= T; t++)
			solve(t);
		out.flush();
		out.close();
	}

	void pre() throws Exception {
	}

	void solve(int TC) throws Exception {
		int n = ni();
		int k = ni();
		int a[] = new int[k + 1];
		for (int i = 1; i <= k; i++) {
			int num = ni();
			a[i] = num;
		}
		long arr[] = new long[n + 1];
		Arrays.fill(arr, Integer.MAX_VALUE);
		for (int i = 1; i <= k; i++) {
			int t = ni();
			arr[a[i]] = t;
		}
		long suffix[] = new long[n + 1];
		long prefix[] = new long[n + 1];
		long cur = Integer.MAX_VALUE -10;
		for (int i = 1; i <= n; i++) {
			cur += 1;
			prefix[i] = Math.min(cur, arr[i]);
			cur = Math.min(cur, arr[i]);
		}
		cur = Integer.MAX_VALUE -10;
		for (int i = n; i > 0; i--) {
			cur += 1;
			suffix[i] = Math.min(cur, arr[i]);
			cur = Math.min(cur, arr[i]);
		}

		for (int i = 1; i <= n; i++) {
			p(Math.min(prefix[i], suffix[i]) + " ");
		}

		p("\n");

	}

	static int bitCount(int x) {
		return x == 0 ? 0 : (1 + bitCount(x & (x - 1)));
	}

	static void dbg(Object... o) {
		System.err.println(Arrays.deepToString(o));
	}

	public static boolean isPrime(long n) {
		if (n < 2)
			return false;
		if (n == 2 || n == 3)
			return true;
		if (n % 2 == 0 || n % 3 == 0)
			return false;
		long sqrtN = (long) Math.sqrt(n) + 1;
		for (long i = 6L; i <= sqrtN; i += 6) {
			if (n % (i - 1) == 0 || n % (i + 1) == 0)
				return false;
		}
		return true;
	}

	public static long gcd(long a, long b) {
		if (a > b)
			a = (a + b) - (b = a);
		if (a == 0L)
			return b;
		return gcd(b % a, a);
	}

	public long pow(long base, long exp, long MOD) {
		base %= MOD;
		long res = 1;
		while (exp > 0) {
			if ((exp & 1) == 1)
				res = res * base % MOD;
			base = base * base % MOD;
			exp >>= 1;
		}
		return res;
	}

	public static long totient(long n) {
		long result = n;
		for (int p = 2; p * p <= n; ++p)
			if (n % p == 0) {
				while (n % p == 0)
					n /= p;
				result -= result / p;
			}
		if (n > 1)
			result -= result / n;
		return result;
	}

	int bit(long n) {
		return (n == 0) ? 0 : (1 + bit(n & (n - 1)));
	}

	public long max(long... arr) {
		long max = arr[0];
		for (long itr : arr)
			max = Math.max(max, itr);
		return max;
	}

	public int max(int... arr) {
		int max = arr[0];
		for (int itr : arr)
			max = Math.max(max, itr);
		return max;
	}

	public long min(long... arr) {
		long min = arr[0];
		for (long itr : arr)
			min = Math.min(min, itr);
		return min;
	}

	public int min(int... arr) {
		int min = arr[0];
		for (int itr : arr)
			min = Math.min(min, itr);
		return min;
	}

	public long sum(long... arr) {
		long sum = 0;
		for (long itr : arr)
			sum += itr;
		return sum;
	}

	public long sum(int... arr) {
		long sum = 0;
		for (int itr : arr)
			sum += itr;
		return sum;
	}

	void p(Object o) {
		out.print(o);
	}

	void pn(Object o) {
		out.println(o);
	}

	void pni(Object o) {
		out.println(o);
		out.flush();
	}

	String n() throws Exception {
		return in.next();
	}

	String nln() throws Exception {
		return in.nextLine();
	}

	int ni() throws Exception {
		return Integer.parseInt(in.next());
	}

	long nl() throws Exception {
		return Long.parseLong(in.next());
	}

	double nd() throws Exception {
		return Double.parseDouble(in.next());
	}

	class FastReader {
		BufferedReader br;
		StringTokenizer st;

		public FastReader() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}

		public FastReader(String s) throws Exception {
			br = new BufferedReader(new FileReader(s));
		}

		String next() throws Exception {
			while (st == null || !st.hasMoreElements()) {
				try {
					st = new StringTokenizer(br.readLine());
				} catch (IOException e) {
					throw new Exception(e.toString());
				}
			}
			return st.nextToken();
		}

		String nextLine() throws Exception {
			String str = "";
			try {
				str = br.readLine();
			} catch (IOException e) {
				throw new Exception(e.toString());
			}
			return str;
		}
	}
}