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 a[20];
int s[20];
long long arr[2005];
long long prv[2005];
long long rev(long long x) {
  long long deg = 1000000007 - 2, ret = 1;
  while (deg) {
    if (deg & 1) {
      ret *= x;
      ret %= 1000000007;
    }
    x *= x;
    x %= 1000000007;
    deg >>= 1;
  }
  return ret;
}
long long C(long long n, long long k) {
  long long ret = 1;
  for (int i = 0; i < k; ++i) {
    ret *= (n - i);
    ret %= 1000000007;
    ret *= rev(i + 1);
    ret %= 1000000007;
  }
  return ret;
}
int main() {
  int n, k;
  cin >> n >> k;
  for (int i = 0; i < n; ++i) {
    cin >> arr[i];
  }
  for (int i = 0; i < n; ++i) {
    prv[i] = C(k + i - 1, i);
  }
  for (int i = 0; i < n; ++i) {
    long long sum = 0;
    for (int j = 0; j <= i; ++j) {
      sum += prv[i - j] * arr[j];
      sum %= 1000000007;
    }
    cout << sum << " ";
  }
  cout << endl;
  return 0;
}
