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 int a[2001];
long long int vals[2001];
long long int val(long long int i) {
  long long int x = 1, y = i, b = 1000000005, c = 1000000007;
  while (b) {
    if (b % 2 == 1) x = x * y % c;
    y = y * y % c;
    b = b / 2;
  }
  return x % c;
}
void comp(long long int n, long long int k) {
  long long int ans = 1, i;
  vals[0] = 1;
  for (i = 1; i < n; i++) {
    ans = ans * (k + i);
    ans %= 1000000007;
    ans = ans * val(i);
    ans %= 1000000007;
    vals[i] = ans;
  }
  return;
}
int main() {
  long long int n, k, i, j;
  cin >> n >> k;
  for (i = 1; i <= n; i++) {
    cin >> a[i];
  }
  long long int ans = 0;
  comp(n, k - 1);
  for (i = n; i >= 1; i--) {
    ans = 0;
    for (j = 1; j <= i; j++) {
      ans = ans + a[j] * vals[i - j];
      ans %= 1000000007;
    }
    a[i] = ans;
  }
  for (i = 1; i <= n; i++) cout << a[i] << ' ';
  printf("\n");
  return 0;
}
