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 int MOD = 1000000007;
vector<long long> product(const vector<long long>& x,
                          const vector<long long>& y) {
  int n = x.size();
  vector<long long> ret(n, 0);
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j <= i; ++j) {
      ret[i] += x[i - j] * y[j];
      ret[i] %= MOD;
    }
  }
  return ret;
}
int main() {
  int n, k;
  cin >> n >> k;
  vector<long long> a(n);
  for (int i = 0; i < n; ++i) cin >> a[i];
  vector<long long> x(n, 1);
  vector<long long> y(n, 0);
  y[0] = 1;
  vector<long long> z = x;
  while (k > 0) {
    if (k & 1) y = product(y, z);
    z = product(z, z);
    k >>= 1;
  }
  vector<long long> ret(n);
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j <= i; ++j) {
      ret[i] += y[i - j] * a[j];
      ret[i] %= MOD;
    }
  }
  cout << ret[0];
  for (int i = 1; i < n; ++i) cout << ' ' << ret[i];
  cout << endl;
  return 0;
}
