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 N, k;
vector<int> mult(vector<int> &a, vector<int> &b) {
  vector<int> r(2100);
  int i, j;
  long long t;
  for (i = 0; i < N; ++i) {
    t = 0;
    for (j = 0; j < N; ++j)
      t += ((a[j]) * 1LL * ((i - j >= 0 ? b[i - j] : 0))) % 1000000007;
    r[i] = t % 1000000007;
  }
  return r;
}
vector<int> pow(vector<int> m, int b) {
  vector<int> r(2100);
  int i;
  for (i = 0; i < N; ++i) r[i] = 0;
  r[0] = 1;
  while (b) {
    if (b & 1) {
      --b;
      r = mult(r, m);
    } else {
      m = mult(m, m);
      b >>= 1;
    }
  }
  return r;
}
int main() {
  cin >> N >> k;
  int i;
  vector<int> m(2100);
  for (i = 0; i < N; ++i) cin >> m[i];
  if (k == 0) {
    for (i = 0; i < N; ++i) cout << m[i] << " ";
    return 0;
  }
  vector<int> mm(2100);
  for (i = 0; i < N; ++i) mm[i] = 1;
  mm = pow(mm, k);
  long long a;
  for (i = 0; i < N; ++i) {
    a = 0;
    for (int j = 0; j < N; ++j)
      a += (m[j] * 1LL * (i - j >= 0 ? mm[i - j] : 0)) % 1000000007;
    cout << a % 1000000007 << " ";
  }
  return 0;
}
