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 M = 1000000007;
long long rev(long long b) {
  long long t = 1;
  for (long long p = M - 2; p > 0; p >>= 1) {
    if (p & 1) (t *= b) %= M;
    (b *= b) %= M;
  }
  return t;
}
int main(void) {
  long long n, k, a[2048], b[2048] = {1};
  cin >> n >> k;
  for (int i = 0; i < n; ++i) cin >> a[i];
  for (int i = 0; i < n; ++i) b[i + 1] = b[i] * (k + i) % M * rev(i + 1) % M;
  for (int i = 0; i < n; ++i) {
    long long s = 0;
    for (int j = 0; j <= i; ++j) (s += a[j] * b[i - j]) %= M;
    cout << s << " ";
  }
  cout << endl;
}
