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.BufferedReader;
import java.io.BufferedWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.util.StringTokenizer;

public class C {

	static StringTokenizer st;
	static BufferedReader in;
	static int p = (int) (1e9+7);
	
	public static void main(String[] args) throws IOException{
		in = new BufferedReader(new InputStreamReader(System.in));
		PrintWriter pw = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
		int n = nextInt();
		int k = nextInt();
		int[]a = new int[n+1];
		for (int i = 1; i <= n; i++) {
			a[i] = nextInt();
		}
		int[]b = new int[n+1];
		b[1] = 1;
		for (int i = 2; i <= n; i++) {
			b[i] = (int) ((long)b[i-1]*(k+i-2) % p * invers_in_modul(i-1) % p);
		}
		int[]ans = new int[n+1];
		for (int i = 1; i <= n; i++) {
			for (int j = 1; j <= i; j++) {
				ans[i] = (ans[i]+(int)((long)b[j]*a[i+1-j] % p)) % p;
			}
		}
		for (int i = 1; i <= n; i++) {
			pw.print(ans[i]+" ");
		}
		pw.close();
	}

	private static int invers_in_modul(int a) {
		int n = p-2;
		int res = 1;
		while (n > 0) {
			if ((n & 1)==1)
				res = (int) ((long)res*a % p);
			n >>= 1;
			a = (int) ((long)a*a % p);
		}
		return res;
	}

	private static int nextInt() throws IOException{
		return Integer.parseInt(next());
	}
	
	private static String next() throws IOException {
        while (st == null || !st.hasMoreTokens()) {
            st = new StringTokenizer(in.readLine());
        }
        return st.nextToken();
    }
}
