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 extgcd(int a, int b, int &x, int &y) {
  int d = a;
  x = 1;
  y = 0;
  if (b) d = extgcd(b, a % b, y, x), y -= a / b * x;
  return d;
}
int modinv(int a) {
  int x, y;
  extgcd(a, MOD, x, y);
  return (x + MOD) % MOD;
}
int main() {
  int n, k, a[2010], b[2010];
  long long c[2010];
  cin >> n >> k;
  for (int i = 0; i < (int)(n); ++i) cin >> a[i];
  if (k == 0)
    copy(a, a + n, b);
  else {
    --k;
    fill_n(b, n, 0);
    c[0] = 1;
    for (int i = 1; i < (int)(n); ++i)
      c[i] = c[i - 1] * (k + i) % MOD * modinv(i) % MOD;
    for (int i = 0; i < (int)(n); ++i)
      for (int j = 0; j < (int)(i + 1); ++j)
        b[i] = (b[i] + a[j] * c[i - j]) % MOD;
  }
  for (int i = 0; i < (int)(n); ++i) cout << b[i] << (i == n - 1 ? "\n" : " ");
}
