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>
long long int arr[2010], s[2010];
long long int inv[2010][35], modinv[2010];
long long int n, k;
int main(void) {
  long long int nck, p;
  long long int i, j;
  scanf("%I64d %I64d", &n, &k);
  for (i = 1; i <= n; i++) {
    scanf("%I64d", &arr[i]);
  }
  if (k == 0) {
    for (i = 1; i <= n; i++) {
      printf("%I64d ", arr[i]);
    }
    return false;
  }
  k--;
  bool f;
  for (i = 1; i <= n; i++) {
    inv[i][0] = i;
    for (j = 1; j <= 30; j++) {
      inv[i][j] = inv[i][j - 1] * inv[i][j - 1];
      inv[i][j] %= 1000000007;
    }
    modinv[i] = 1;
    for (j = 0; j <= 30; j++) {
      if ((1000000005 >> j) & 1) {
        modinv[i] *= inv[i][j];
        modinv[i] %= 1000000007;
      }
    }
  }
  for (i = 1; i <= n; i++) {
    nck = 1;
    p = 0;
    f = 0;
    for (j = i; j >= 1; j--) {
      s[i] += nck * arr[j];
      nck %= 1000000007;
      p++;
      if (f == 1) {
        nck *= (k + p);
        nck %= 1000000007;
        nck *= modinv[p];
        nck %= 1000000007;
      } else {
        nck *= (k + p);
        nck /= p;
        if (nck >= 1000000007) {
          f = 1;
          nck %= 1000000007;
        }
      }
      s[i] %= 1000000007;
    }
  }
  for (i = 1; i <= n; i++) {
    printf("%I64d ", s[i]);
  }
  return false;
}
