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 n, k, i, j, b[2002], l, s, mod = 1000000007;
long long a[2002], arr[500000], sum;
int gcd(int x, int y) {
  if (x % y == 0) return y;
  return gcd(y, x % y);
}
int main() {
  scanf("%d%d", &n, &k);
  b[0] = k;
  for (i = 1; i < n; i++) b[i] = b[i - 1] + 1;
  a[0] = 1;
  for (i = 1; i < n; i++) {
    for (l = i, j = 0, sum = 1; j < i; j++) {
      if (l > 1) {
        s = gcd(b[j], l);
        b[j] /= s;
        l /= s;
      }
      sum = (sum * b[j]) % mod;
    }
    a[i] = sum;
  }
  for (i = 0; i < n; i++) scanf("%I64d", &arr[i]);
  printf("%I64d ", arr[0]);
  for (i = 1; i < n; i++) {
    for (sum = j = 0; j <= i; j++) sum = (sum + arr[j] * a[i - j]) % mod;
    printf("%I64d ", sum);
  }
  scanf(" ");
}
