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 Solution
{
    static class Reader {
        final private int BUFFER_SIZE = 1 << 16;
        private DataInputStream din;
        private byte[] buffer;
        private int bufferPointer, bytesRead;
 
        public Reader()
        {
            din = new DataInputStream(System.in);
            buffer = new byte[BUFFER_SIZE];
            bufferPointer = bytesRead = 0;
        }
 
        public Reader(String file_name) throws IOException
        {
            din = new DataInputStream(
                new FileInputStream(file_name));
            buffer = new byte[BUFFER_SIZE];
            bufferPointer = bytesRead = 0;
        }
 
        public String readLine() throws IOException
        {
            byte[] buf = new byte[64]; // line length
            int cnt = 0, c;
            while ((c = read()) != -1) {
                if (c == '\n') {
                    if (cnt != 0) {
                        break;
                    }
                    else {
                        continue;
                    }
                }
                buf[cnt++] = (byte)c;
            }
            return new String(buf, 0, cnt);
        }
 
        public int nextInt() throws IOException
        {
            int ret = 0;
            byte c = read();
            while (c <= ' ') {
                c = read();
            }
            boolean neg = (c == '-');
            if (neg)
                c = read();
            do {
                ret = ret * 10 + c - '0';
            } while ((c = read()) >= '0' && c <= '9');
 
            if (neg)
                return -ret;
            return ret;
        }
 
        public long nextLong() throws IOException
        {
            long ret = 0;
            byte c = read();
            while (c <= ' ')
                c = read();
            boolean neg = (c == '-');
            if (neg)
                c = read();
            do {
                ret = ret * 10 + c - '0';
            } while ((c = read()) >= '0' && c <= '9');
            if (neg)
                return -ret;
            return ret;
        }
 
        public double nextDouble() throws IOException
        {
            double ret = 0, div = 1;
            byte c = read();
            while (c <= ' ')
                c = read();
            boolean neg = (c == '-');
            if (neg)
                c = read();
 
            do {
                ret = ret * 10 + c - '0';
            } while ((c = read()) >= '0' && c <= '9');
 
            if (c == '.') {
                while ((c = read()) >= '0' && c <= '9') {
                    ret += (c - '0') / (div *= 10);
                }
            }
 
            if (neg)
                return -ret;
            return ret;
        }
 
        private void fillBuffer() throws IOException
        {
            bytesRead = din.read(buffer, bufferPointer = 0,
                                 BUFFER_SIZE);
            if (bytesRead == -1)
                buffer[0] = -1;
        }
 
        private byte read() throws IOException
        {
            if (bufferPointer == bytesRead)
                fillBuffer();
            return buffer[bufferPointer++];
        }
 
        public void close() throws IOException
        {
            if (din == null)
                return;
            din.close();
        }
    }
    
   
  static int parent(int a , int p[])
  {
      if(a == p[a])
      return a;
      
      return p[a] = parent(p[a],p);
  }
  
  static void union(int a , int b , int p[] , int size[])
  {
      a = parent(a,p);
      b = parent(b,p);
      
      if(a == b)
      return;
      
      if(size[a] < size[b])
      {
          int temp = a;
          a = b;
          b = temp;
      }
      
      p[b] = a;
      size[a] += size[b];
  }
  
  
  static long dp[][];
  static long f(int n , int m)
  {
      if(m == 0)
      return 1;
      
      if(dp[n][m] != -1)
      return dp[n][m];
      
      if(n < 9)
      dp[n][m] = f(n+1,m-1);
      
      else
      dp[n][m] = f(0,m-1)+f(1,m-1);
      
      dp[n][m] %= 1000000007;
      
      return dp[n][m];
  }
  
 static long getSum(int index , long BITree[])
    {
        long sum = 0; // Iniialize result
      
        // index in BITree[] is 1 more than
        // the index in arr[]
       // index = index + 1;
      
        // Traverse ancestors of BITree[index]
        while(index>0)
        {
            // Add current element of BITree 
            // to sum
            sum += BITree[index];
      
            // Move index to parent node in 
            // getSum View
            index -= index & (-index);
        }
        return sum;
    }
  
    // Updates a node in Binary Index Tree (BITree)
    // at given index in BITree. The given value 
    // 'val' is added to BITree[i] and all of 
    // its ancestors in tree.
    public static void updateBIT(int n, int index, 
                                        long val , long BITree[])
    {
        // index in BITree[] is 1 more than 
        // the index in arr[]
      //  index = index + 1;
      
        // Traverse all ancestors and add 'val'
        while(index <= n)
        {
        // Add 'val' to current node of BIT Tree
        BITree[index] += val;
      
        // Update index to that of parent 
        // in update View
        index += index & (-index);
        }
    }
    
    public static void main(String []args) throws IOException
    {
        Reader sc = new Reader();
        int t = sc.nextInt();
        while(t-- > 0)
        {
            int n = sc.nextInt();
            int k = sc.nextInt();
            int pos[] = new int[k];
            for(int i = 0 ; i < k ; i++)
            {
                pos[i] = sc.nextInt()-1;
            }
            int temp[] = new int[n];
            TreeMap<Integer,Integer> map1 = new TreeMap<Integer,Integer>();
            TreeMap<Integer,Integer> map2  = new TreeMap<Integer,Integer>();
            for(int i = 0 ; i < k ; i++)
            {
                int x = sc.nextInt();
                temp[pos[i]] = x;
                
                if(map2.containsKey(x+pos[i]+1))
                map2.replace(x+pos[i]+1,map2.get(x+pos[i]+1)+1);
                
                else
                map2.put(x+pos[i]+1,1);
            }
            StringBuffer str = new StringBuffer("");
            
            for(int i = 0 ; i < n ; i++)
            {
                if(temp[i] != 0)
                {
                     if(map2.get(temp[i]+i+1) == 1)
                     map2.remove(temp[i]+i+1);
                     
                     else
                     map2.replace(temp[i]+i+1,map2.get(temp[i]+i+1)-1);
                     
                     if(map1.containsKey(temp[i]-i-1))
                     map1.replace(temp[i]-i-1 , map1.get(temp[i]-i-1)+1);
                     
                     else
                     map1.put(temp[i]-i-1 , 1);
                }
                
                int min = Integer.MAX_VALUE;
                if(map1.size() > 0)
                min = Math.min(min,map1.firstKey()+i+1);
                
                if(map2.size() > 0)
                min = Math.min(min,map2.firstKey()-i-1);
                
                str.append(min+ " ");
            }
            System.out.println(str);
        }
    }
}