You've got an array a, consisting of n integers. The array elements are indexed from 1 to n. Let's determine a two step operation like that:

  1. First we build by the array a an array s of partial sums, consisting of n elements. Element number i (1 ≤ i ≤ n) of array s equals <image>. The operation x mod y means that we take the remainder of the division of number x by number y. 
  2. Then we write the contents of the array s to the array a. Element number i (1 ≤ i ≤ n) of the array s becomes the i-th element of the array a (ai = si). 



You task is to find array a after exactly k described operations are applied.

Input

The first line contains two space-separated integers n and k (1 ≤ n ≤ 2000, 0 ≤ k ≤ 109). The next line contains n space-separated integers a1, a2, ..., an — elements of the array a (0 ≤ ai ≤ 109).

Output

Print n integers — elements of the array a after the operations are applied to it. Print the elements in the order of increasing of their indexes in the array a. Separate the printed numbers by spaces.

Examples

Input

3 1
1 2 3


Output

1 3 6


Input

5 0
3 14 15 92 6


Output

3 14 15 92 6
import java.math.BigInteger;
import java.util.Scanner;


public class Main {

	/**
	 * @param args
	 */
	public static void main(String[] args) {
		// TODO Auto-generated method stub
			Scanner cin=new Scanner(System.in);
			BigInteger tab[]=new BigInteger[2004];
			BigInteger  res;
			int n;
			int i,j,k;
			int num[]=new int[2004];
			n=cin.nextInt();
			k=cin.nextInt();
			for(i=1;i<=n;i++){
				num[i]=cin.nextInt();
			}
			tab[0]=BigInteger.ONE;
			for(i=1;i<=n;i++)
				tab[i]=tab[i-1]
						.multiply(BigInteger.valueOf(k+i-1)).divide(BigInteger.valueOf(i));
			for(i=1;i<=n;i++)
				tab[i]=tab[i].mod(BigInteger.valueOf(1000000007));
			for(i=1;i<=n;i++){
				res=BigInteger.ZERO;
				for(j=0;j<i;j++){
					res=res.add(BigInteger.valueOf(num[i-j]).multiply(tab[j]))
							.mod(BigInteger.valueOf(1000000007));
				}
				if(i==n)System.out.println(res);
				else System.out.print(res+" ");
			}
	}
}

  			  					 	 	  						     	