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 num = 1e9 + 7, n, coff[2100], ans[2100], temp[2100];
void pwr(long long k) {
  int i, j;
  for (i = 0; i < n; i++) coff[i] = 1, ans[i] = 1;
  while (k > 0) {
    if (k & 1) {
      for (i = 0; i < n; i++) ans[i] = coff[i];
    }
    for (i = 0; i < n; i++) {
      temp[i] = 0;
      for (j = i; j >= 0; j--)
        temp[i] = (temp[i] + coff[j] * coff[i - j]) % num;
    }
    for (i = 0; i < n; i++) coff[i] = temp[i];
    k >>= 1;
  }
}
int main() {
  int i, j;
  long long k, a[2100];
  scanf("%I64d %I64d", &n, &k);
  for (i = 0; i < n; i++) scanf("%I64d", &a[i]);
  pwr(k);
  if (k == 0) {
    for (i = 0; i < n; i++) {
      printf("%I64d ", a[i]);
    }
  } else
    for (i = 0; i < n; i++) {
      long long t1 = 0;
      for (j = i; j >= 0; j--) t1 = (t1 + ans[j] * a[i - j]) % num;
      printf("%I64d ", t1);
    }
  printf("\n");
}
