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;
long long mult[3000];
int init[3000];
long long ans[3000];
int put(int x, int p);
int inv_mod(int x);
int main() {
  int n, k;
  cin >> n >> k;
  k--;
  mult[0] = 1;
  for (int i(1); i < 3000; i++) {
    mult[i] = (mult[i - 1] * (k + i)) % 1000000007ll;
    mult[i] = (mult[i] * inv_mod(i)) % 1000000007ll;
  }
  int c;
  for (int i(1); i <= n; i++) {
    cin >> c;
    for (int j(i); j <= n; j++) {
      long long r((mult[j - i] * c) % 1000000007ll);
      ans[j] = (ans[j] + r) % 1000000007ll;
    }
  }
  for (int i(1); i <= n; i++) cout << ans[i] << ' ';
  return 0;
}
int inv_mod(int x) { return put(x, 1000000007ll - 2); }
int put(int x, int p) {
  if (p == 0) return 1;
  long long a(put(x, p / 2));
  a = (a * a) % 1000000007ll;
  if ((p & 1)) a = (a * x) % 1000000007ll;
  return a;
}
