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 a[100100], b[100100], c[100100];
long long expmod(long long a, long long b) {
  long long rlt = 1, x = a;
  while (b) {
    if (b & 1) rlt = rlt * x % 1000000007;
    x = x * x % 1000000007, b >>= 1;
  }
  return rlt;
}
int main() {
  scanf("%d %d", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  if (k == 0) {
    for (int i = 1; i <= n; i++) cout << a[i] << ' ';
    cout << endl;
    return 0;
  }
  c[1] = 1;
  for (int i = 2; i <= n; i++) {
    c[i] = (c[i - 1] * (k + i - 2) % 1000000007) *
           expmod(i - 1, 1000000007 - 2) % 1000000007;
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= i; j++)
      b[i] = (b[i] + c[i - j + 1] * a[j]) % 1000000007;
  }
  for (int i = 1; i <= n; i++) cout << b[i] << ' ';
  cout << endl;
  return 0;
}
