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 a[2010], sum;
long long num[2010];
long long c(int a, int b) {
  int t = a, ans = 1;
  while (b) {
    if (b & 1) ans = (long long)ans * t % 1000000007;
    b >>= 1;
    t = (long long)t * t % 1000000007;
  }
  return ans;
}
int main() {
  int n, k;
  long long tmp;
  while (~scanf("%d%d", &n, &k)) {
    for (int i = 0; i < n; i++) {
      scanf("%I64d", &a[i]);
    }
    if (k == 0) {
      for (int i = 0; i < n; i++) cout << a[i] << " ";
      cout << endl;
    } else {
      sum = a[0];
      num[0] = 1;
      cout << sum << " ";
      for (int i = 1; i < n; i++) {
        sum = 0;
        num[i] = num[i - 1] * c(i, 1000000005) % 1000000007 * (k + i - 1) %
                 1000000007;
        for (int j = 0; j <= i; j++) {
          sum = (long long)(sum + num[j] * a[i - j]) % 1000000007;
        }
        cout << sum << " ";
      }
      cout << endl;
    }
  }
  return 0;
}
