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>
#pragma GCC optimize("Ofast,no-stack-protector")
using namespace std;
using namespace rel_ops;
using ll = long long;
using Vi = vector<int>;
using Pii = pair<int, int>;
struct DD {
  using z = void;
  template <class t, class... v>
  DD& operator,(t first) {
    p(first, 0);
    return *this;
  }
  template <class t, class... v>
  z p(t first, ...) {
    cerr << first;
  }
  template <class t, class... v>
  auto p(t first, int) -> decltype(&t::print, z()) {
    DD();
    first.print();
  }
  template <class t, class... v>
  auto p(t first, int) -> decltype(t().first, z()) {
    DD(), "(", first.first, ", ", first.second, ")";
  }
  template <class t, class... v>
  auto p(t first, int) -> decltype(t().begin(), z()) {
    DD(), "[";
    for (auto& e : (first)) DD(), e, ", ";
    DD(), "]";
  }
  z d(...) {}
  template <class t, class... v>
  z d(const char* s, t a, v... b) {
    while (*s && *s != ',') DD(), *s++;
    DD(), ": ", a, *s++;
    d(s, b...);
  }
};
vector<int> mc(vector<int>& a, const vector<int>& b) {
  vector<int> c = a;
  int n = int((a).size());
  for (int i = 0; i < n; ++i) {
    for (int j = 1; j <= i; ++j) {
      c[i] += a[i - j] * b[j];
    }
  }
  return c;
}
void power(vector<int>& a, int e, int n) {
  vector<int> b(n, 1);
  while (e) {
    if (e & 1) a = mc(a, b);
    e >>= 1;
    b = mc(b, b);
  }
}
int main() {
  cin.sync_with_stdio(0);
  cin.tie(0);
  cin.exceptions(cin.failbit);
  cout << fixed << setprecision(10);
  int n, k;
  cin >> n >> k;
  vector<int> a(n);
  for (auto& i : (a)) cin >> i;
  power(a, k, n);
  for (auto& i : (a)) cout << i << ' ';
  cout << endl;
}
