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;
int n;
vector<int> multiply(const vector<int>& a, const vector<int>& b) {
  vector<int> res(n);
  for (int i = 0; i < n; i++)
    for (int j = 0; i + j < n; j++)
      res[i + j] = (res[i + j] + a[i] * 1LL * b[j]) % MOD;
  return res;
}
vector<int> pow(vector<int> a, int st) {
  vector<int> res(n, 0);
  res[0] = 1;
  for (int i = 0; (1 << i) <= st; i++) {
    if (st & (1 << i)) res = multiply(res, a);
    a = multiply(a, a);
  }
  return res;
}
int main() {
  scanf("%d", &n);
  int k;
  scanf("%d", &k);
  vector<int> a(n);
  vector<int> b(n);
  for (int i = 0; i < n; i++) scanf("%d", &a[i]), b[i] = 1;
  b = pow(b, k);
  a = multiply(a, b);
  for (int i = 0; i < n; i++) printf("%d ", a[i]);
  printf("\n");
  return 0;
}
