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.Arrays;
import java.util.Scanner;
public class E {
	static long M = 1000*1000*1000+7;
	static long[] f;
	static long[] finv;
	static long K;
	
	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		int n = in.nextInt();
		int k = in.nextInt();
		K = k;
		long[] A = new long[n];
		for(int i=0; i<n; i++)
			A[i] = in.nextInt();
		
		f = new long[n];
		finv = new long[n];
		f[0] = 1;
		finv[0] = 1;
		for(int i=1; i<n; i++) {
			f[i] = (f[i-1]*i)%M;
			finv[i] = minv(f[i]);
		}
		
		long[] choose = new long[n];
		for(int i=1; i<n; i++) {
			choose[i] = 1;
			for(long j=k; j<=k+i-1; j++) {
				choose[i] = (choose[i]*j)%M;
			}
			choose[i] = (choose[i]*finv[i])%M;	
		}
		
		
		long[] B = new long[n];
		for(int i=0; i<n; i++) {
			for(int j=0; j<i; j++) {
				B[i] = (B[i] + ((choose[i-j]*A[j])%M))%M;
			}
			B[i] = (B[i]+A[i])%M;
		}
		for(int i=0; i<n; i++) {
			System.out.print(B[i]+(i==n-1?"\n":" "));
		}
	}
	
	static BigInteger MM = BigInteger.valueOf(M);
	static long minv(long x) {
		return BigInteger.valueOf(x).modInverse(MM).longValue();
	}
}
