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.File;
import java.io.FileReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.math.BigInteger;
import java.util.StringTokenizer;

public class Solution implements Runnable {
	private BufferedReader br;
	private StringTokenizer tok;
	private PrintWriter out;
	
	static final boolean ONLINE_JUDGE = System.getProperty("ONLINE_JUDGE") != null;
	
	static final int MOD = 1000 * 1000 * 1000 + 9;
			
	void solve() throws IOException {
		int n = nextInt();
		int k = nextInt() - 1;
		int[] a = new int[n];
		for (int i = 0; i < n; ++i) {
			a[i] = nextInt();
		}
		if (k < 0) {
			for (int i = 0; i < n; ++i) {
				out.print(a[i] + " ");
			}
			return;
		}
		long[] c = new long[n];
		c[0] = 1;
		for (int i = 1; i < n; ++i) {
			c[i] = c[i - 1] * (k + i);
			c[i] %= MOD;
			c[i] *= BigInteger.valueOf(i).modInverse(BigInteger.valueOf(MOD)).longValue();
			c[i] %= MOD;
		}
		long[] res = new long[n];
		for (int i = 0; i < n; ++i) {
			for (int j = i; j < n; ++j) {
				res[j] += c[j - i] * a[i] % MOD;
				res[j] %= MOD;
			}
		}
		for (int i = 0; i < n; ++i) {
			out.print(res[i] + " ");
		}
	}
		
	public void run() {
		try {
			if (ONLINE_JUDGE) {
				br = new BufferedReader(new InputStreamReader(System.in));
				out = new PrintWriter(System.out);
			} else {
				br = new BufferedReader(new FileReader(new File("input.txt")));
				out = new PrintWriter(new File("output.txt"));
			}
			solve();
			br.close();
			out.close();
		} catch (IOException e) {
			e.printStackTrace();
			System.exit(1);
		}
	}
	
	public static void main(String[] args) {
		new Solution().run();
	}	
	
	String nextToken() throws IOException {
		while (tok == null || !tok.hasMoreTokens())
			tok = new StringTokenizer(br.readLine());
		return tok.nextToken();
	}
	
	String nextString() throws IOException {
		return nextToken();
	}
	
	int nextInt() throws IOException {
		return Integer.parseInt(nextToken());
	}
	
	long nextLong() throws IOException {
		return Long.parseLong(nextToken());
	}
	
	double nextDouble() throws IOException {
		return Double.parseDouble(nextToken());
	}
	
	BigInteger nextBigInteger() throws IOException {
		return new BigInteger(nextToken());
	}
}