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>
using namespace std;
long long Polibroj[2005], N, K;
long long A[2005], R[2005];
void ucitaj() {
  scanf("%I64d%I64d", &N, &K);
  int i;
  for (i = 1; i <= N; i++) scanf("%I64d", A + i);
}
long long stepen(long long x, long long e) {
  if (e == 0) return 1;
  if (e % 2 == 0) return (stepen(x, e / 2) * stepen(x, e / 2)) % (1000000007ll);
  return (((stepen(x, e / 2) * stepen(x, e / 2)) % (1000000007ll)) * x) %
         (1000000007ll);
}
long long inv(long long x) { return stepen(x, (1000000007ll) - 2); }
void napravi() {
  Polibroj[1] = 1;
  Polibroj[2] = K;
  long long i;
  for (i = 3; i <= N; i++) {
    Polibroj[i] =
        (((Polibroj[i - 1] * (K + i - 2)) % (1000000007ll)) * inv(i - 1)) %
        (1000000007ll);
  }
}
void resi() {
  long long i, j, k;
  for (i = 1; i <= N; i++) {
    for (j = 1; j <= i; j++) {
      R[i] += A[j] * Polibroj[i + 1 - j];
      R[i] %= (1000000007ll);
    }
  }
  for (i = 1; i <= N; i++) printf("%I64d ", R[i]);
  puts("");
}
int main() {
  ucitaj();
  napravi();
  resi();
  return 0;
}
