Problem: Artem is building a new robot. He has a matrix a consisting of n rows and m columns. The cell located on the i-th row from the top and the j-th column from the left has a value a_{i,j} written in it. 

If two adjacent cells contain the same value, the robot will break. A matrix is called good if no two adjacent cells contain the same value, where two cells are called adjacent if they share a side. 

Artem wants to increment the values in some cells by one to make a good.

More formally, find a good matrix b that satisfies the following condition — 

  * For all valid (i,j), either b_{i,j} = a_{i,j} or b_{i,j} = a_{i,j}+1. 



For the constraints of this problem, it can be shown that such a matrix b always exists. If there are several such tables, you can output any of them. Please note that you do not have to minimize the number of increments.

Input

Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10). Description of the test cases follows.

The first line of each test case contains two integers n, m (1 ≤ n ≤ 100, 1 ≤ m ≤ 100) — the number of rows and columns, respectively.

The following n lines each contain m integers. The j-th integer in the i-th line is a_{i,j} (1 ≤ a_{i,j} ≤ 10^9).

Output

For each case, output n lines each containing m integers. The j-th integer in the i-th line is b_{i,j}.

Example

Input


3
3 2
1 2
4 5
7 8
2 2
1 1
3 3
2 2
1 3
2 2


Output


1 2
5 6
7 8
2 1
4 3
2 4
3 2

Note

In all the cases, you can verify that no two adjacent cells have the same value and that b is the same as a with some values incremented by one. 

Solution (in JAVA):

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

public class C {
    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

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

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

        int nextInt() {
            return Integer.parseInt(next());
        }

        long nextLong() {
            return Long.parseLong(next());
        }

        double nextDouble() {
            return Double.parseDouble(next());
        }

        String nextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }

    static FastReader s = new FastReader();
    static PrintWriter out = new PrintWriter(System.out);

    private static int[] rai(int n) {
        int[] arr = new int[n];
        for (int i = 0; i < n; i++) {
            arr[i] = s.nextInt();
        }
        return arr;
    }

    private static int[][] rai(int n, int m) {
        int[][] arr = new int[n][m];
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                arr[i][j] = s.nextInt();
            }
        }
        return arr;
    }

    private static long[] ral(int n) {
        long[] arr = new long[n];
        for (int i = 0; i < n; i++) {
            arr[i] = s.nextLong();
        }
        return arr;
    }

    private static long[][] ral(int n, int m) {
        long[][] arr = new long[n][m];
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                arr[i][j] = s.nextLong();
            }
        }
        return arr;
    }

    private static int ri() {
        return s.nextInt();
    }

    private static long rl() {
        return s.nextLong();
    }

    private static String rs() {
        return s.next();
    }

    static int gcd(int a,int b)
    {
        if(b==0)
        {
            return a;
        }
        return gcd(b,a%b);
    }

    static boolean isPrime(int n) {
        //check if n is a multiple of 2
        if (n % 2 == 0) return false;
        //if not, then just check the odds
        for (int i = 3; i <= Math.sqrt(n); i += 2) {
            if (n % i == 0)
                return false;
        }
        return true;
    }
    static int MOD=1000000007;



    public static HashMap<Long,Integer> primeFactors(long n)
    {
        // Print the number of 2s that divide n
        HashMap<Long,Integer> map=new HashMap<>();

        while (n % 2 == 0) {
            map.put(2L,map.getOrDefault(2L,0)+1);
            n /= 2;
        }

        // n must be odd at this point.  So we can
        // skip one element (Note i = i +2)
        for (long i = 3; i <= Math.sqrt(n); i += 2) {
            // While i divides n, print i and divide n
            while (n % i == 0) {
                map.put(i,map.getOrDefault(i,0)+1);
                n /= i;
            }
        }

        // This condition is to handle the case whien
        // n is a prime number greater than 2
        if (n > 2)
            map.put(n,map.getOrDefault(n,0)+1);

        return map;
    }



    public static void main(String[] args) {
        StringBuilder ans = new StringBuilder();
        int t = ri();
//        int t = 1;
        while (t-- > 0){
            int n=ri();
            int m=ri();
            long[][] arr=ral(n,m);
            boolean[][] vis=new boolean[n][m];
            for(int i=0;i<n;i++)
            {

                for(int j=0;j<m;j++)
                {
                    if(((arr[i][j] ^ i ^ j)&1)!=0)
                    {
                        arr[i][j]++;
                    }
                    ans.append(arr[i][j]).append(" ");
                }
                ans.append("\n");
            }

        }

        out.print(ans.toString());
        out.flush();

    }

}