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;
const unsigned int maxn = 2000;
constexpr unsigned long long mod = 1e9 + 7;
class Number {
 public:
  Number() = default;
  Number(const unsigned long long v) : val(v % mod) {}
  friend inline Number operator+(const Number l, const Number r) {
    return (l.val + mod) + r.val;
  }
  inline Number operator+=(const Number r) { return *this = *this + r; };
  friend inline Number operator-(const Number l, const Number r) {
    return (l.val + mod) - r.val;
  }
  inline Number operator-=(const Number r) { return *this = *this - r; };
  friend inline Number operator*(const Number l, const Number r) {
    return (l.val + mod) * r.val;
  }
  inline Number operator*=(const Number r) { return *this = *this * r; };
  friend inline ostream& operator<<(ostream& os, const Number v) {
    return os << v.val;
  }

 private:
  unsigned long long val = 0;
};
Number quickPow(Number a, unsigned long long e) {
  Number ret = 1;
  for (; e; e >>= 1) {
    if (e & 0x01) ret *= a;
    a *= a;
  }
  return ret;
}
inline Number operator/(const Number a, const Number b) {
  return a * quickPow(b, mod - 2);
}
unsigned int a[maxn + 10];
Number b[maxn + 10], s[maxn + 10];
static void initB(const unsigned int n, const unsigned int k) {
  b[0] = 1;
  for (unsigned int i = 1; i <= n; ++i) b[i] = b[i - 1] * (k + i - 1) / i;
}
static void getAns(const unsigned int n) {
  for (unsigned int i = 0; i < n; ++i)
    for (unsigned int j = 0; j <= i; ++j) s[i] += a[i - j] * b[j];
}
int main() {
  ios_base::sync_with_stdio(false);
  unsigned int n, k;
  cin >> n >> k;
  copy_n(istream_iterator<unsigned int>(cin), n, a);
  initB(n, k);
  getAns(n);
  copy(s, s + n, ostream_iterator<Number>(cout, " "));
  cout << "\n";
  return 0;
}
