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;
int i, j, m, n, p, k, w[2005], a[2005], ans[2005];
int power(int x, int y) {
  int sum = 1;
  for (; y; y >>= 1) {
    if (y & 1) sum = 1ll * sum * x % 1000000007;
    x = 1ll * x * x % 1000000007;
  }
  return sum;
}
int main() {
  w[0] = 1;
  scanf("%d%d", &n, &k);
  for (i = 1; i <= n; ++i) scanf("%d", &a[i]);
  if (k == 0) {
    for (i = 1; i <= n; ++i) printf("%d ", a[i]);
    puts("");
    return 0;
  }
  --k;
  for (i = 1; i <= n; ++i)
    w[i] = 1ll * w[i - 1] * (k + i) % 1000000007 * power(i, 1000000007 - 2) %
           1000000007;
  for (i = 1; i <= n; ++i) {
    for (j = i; j <= n; ++j)
      ans[j] = (ans[j] + 1ll * a[i] * w[j - i] % 1000000007) % 1000000007;
  }
  for (i = 1; i <= n; ++i) printf("%d ", ans[i]);
}
