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
#include <bits/stdc++.h>
int mul(int x, int y) { return (((long long)x) * y) % 1000000007; }
int pw(int c, int x) {
  if (!x) return 1;
  if (x % 2) return mul(c, pw(c, x - 1));
  int b = pw(c, x / 2);
  return mul(b, b);
}
int inv(int c) { return pw(c, 1000000007 - 2); }
int n, k;
int a[2048];
int w[2048];
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 0; i < n; ++i) {
    scanf("%d", a + i);
  }
  w[0] = 1;
  for (int i = 1; i < n; ++i) {
    w[i] = mul(mul(w[i - 1], k + i - 1), inv(i));
  }
  for (int i = 0; i < n; ++i) {
    int s = 0;
    for (int j = 0; j <= i; ++j) {
      s += w[i - j] * a[j];
    }
    if (i) putchar(' ');
    printf("%d", s);
  }
  puts("");
  return 0;
}
