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
// practice with rainboy
import java.io.*;
import java.util.*;

public class CF223C extends PrintWriter {
	CF223C() { super(System.out, true); }
	Scanner sc = new Scanner(System.in);
	public static void main(String[] $) {
		CF223C o = new CF223C(); o.main(); o.flush();
	}

	static final int MD = 1000000007;
	int d_, x_, y_;
	void gcd_(int a, int b) {
		if (b == 0) {
			d_ = a; x_ = 1; y_ = 0;
		} else {
			gcd_(b, a % b);
			int t = x_ - a / b * y_; x_ = y_; y_ = t;
		}
	}
	int inv(int a) {
		gcd_(a, MD);
		return x_ < 0 ? x_ + MD : x_;
	}
	void main() {
		int n = sc.nextInt();
		int k = sc.nextInt();
		int[] aa = new int[n];
		for (int i = 0; i < n; i++)
			aa[i] = sc.nextInt();
		int[] ch = new int[n];
		ch[0] = 1;
		for (int i = 1; i < n; i++)
			ch[i] = (int) ((long) ch[i - 1] * (k - 1 + i) % MD * inv(i) % MD);
		for (int j = n - 1; j >= 0; j--) {
			long a = 0;
			for (int i = 0; i <= j; i++)
				a = (a + (long) aa[i] * ch[j - i]) % MD;
			aa[j] = (int) a;
		}
		for (int i = 0; i < n; i++)
			print(aa[i] + " ");
		println();
	}
}
