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.io.*;
import java.math.BigInteger;
import java.util.*;

public class Codeforces {
	final static long mod = ((long)1e9 + 7);
	final static BigInteger bimod = BigInteger.valueOf(mod);
	
	static long inv(long x) {
		if (x % mod == 0) {
			return 0;
		}
		return BigInteger.valueOf(x).modInverse(bimod).longValue();
	}

	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		int n = in.nextInt();
		long k = in.nextInt();
		long[] a = new long[n];
		for (int i = 0; i < n; i++) {
			a[i] = in.nextInt();
		}
		long[] ainv = new long[n + 1];
		for (int j = 0; j <= n; j++) {
			ainv[j] = inv(j);
		}
		for (int i = 0; i < n; i++) {
			long curc = 1;
			long ans = 0;
			for (int j = 0; j <= i; j++) {
				ans = (ans + a[i - j] * curc) % mod;
				
				if (j != i) {
					curc = (curc * (j + k)) % mod;
					curc = (curc * ainv[j + 1]) % mod;
				}
			}
			System.out.print(ans + " ");
		}

	}
}