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 n, k, inv[2005], C[2005], ans[2005], num[2005];
pair<long long, long long> ex_euclid(long long a, long long b) {
  if (b == 0) {
    return {1, 0};
  }
  pair<long long, long long> tmp = ex_euclid(b, a % b);
  return {tmp.second, tmp.first - (a / b) * tmp.second};
}
long long getinv(long long x) {
  pair<long long, long long> t = ex_euclid(x, 1000000007);
  return (t.first + 1000000007) % 1000000007;
}
void gettb() {
  for (int i = 1; i <= n; i++) {
    inv[i] = getinv(i);
  }
  C[0] = 1;
  for (int i = 1; i <= n; i++) {
    C[i] = (C[i - 1] * (i + k - 1) % 1000000007) * inv[i] % 1000000007;
  }
}
int main() {
  cin >> n >> k;
  gettb();
  for (int i = 1; i <= n; i++) cin >> num[i];
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= i; j++) {
      ans[i] = (ans[i] + (C[i - j] * num[j] % 1000000007)) % 1000000007;
    }
  }
  for (int i = 1; i <= n; i++) cout << ans[i] << ' ';
  return 0;
}
