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 a[2000], b[2000];
int n, k;
int t[2000];
long long inverse(long long a, long long mod) {
  long long b = mod, x = 0, y = 1, t, q, lastx = 1, lasty = 0;
  while (b) {
    q = a / b;
    t = a, a = b, b = t % b;
    t = x, x = lastx - q * x, lastx = t;
    t = y, y = lasty - q * y, lasty = t;
  }
  return a == 1 ? (lastx % mod + mod) % mod : 0;
}
int main() {
  int i, j, mul, div, val;
  scanf("%d %d", &n, &k);
  for (i = 0; i < n; i++) scanf("%d", &a[i]), b[i] = a[i];
  if (!k) goto end;
  for (i = 1; i < n; i++) t[i] = 0;
  t[0] = 1;
  for (i = 0; i < n; i++) {
    for (mul = k, div = 1, val = 1, j = i - 1; j >= 0; j--, mul++, div++) {
      val = ((long long)val * mul) % 1000000007;
      val = (val * inverse(div, 1000000007)) % 1000000007;
      b[i] = (b[i] + (long long)a[j] * val) % 1000000007;
    }
  }
end:
  printf("%d", b[0]);
  for (i = 1; i < n; i++) printf(" %d", b[i]);
  putchar('\n');
  return 0;
}
