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;
int n, k;
long long prod(long long a, long long b) {
  return ((a % 1000000007) * (b % 1000000007)) % 1000000007;
}
long long sum(long long a, long long b) { return (a + b) % 1000000007; }
long long step(long long x, long long n) {
  long long a = 1;
  while (n) {
    if (n & 1) a = prod(a, x);
    n >>= 1;
    x = prod(x, x);
  }
  return a;
}
long long obr(long long x) { return step(x, 1000000007 - 2); }
long long st[2001];
long long pw[2001];
int main() {
  cin >> n >> k;
  int a[2000];
  for (int i = 0; i < n; i++) cin >> a[i];
  st[0] = k;
  pw[0] = 1;
  for (int i = 1; i <= n; i++) {
    st[i] = prod(st[i - 1], k + i);
    pw[i] = prod(pw[i - 1], i);
  }
  for (int i = 2; i <= n; i++) {
    pw[i] = obr(pw[i]);
  }
  long long b[2001];
  for (int i = 0; i < n; i++) {
    b[i] = a[i];
    for (int j = 0; j < i; j++) {
      b[i] = sum(b[i], prod(a[j], prod(st[i - j - 1], pw[i - j])));
    }
  }
  for (int i = 0; i < n; i++) {
    cout << b[i] << " ";
  }
  cout << endl;
  return 0;
}
