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 N = 2005;
const long long mod = 1e9 + 7;
int a[N];
long long val[N];
long long ans[N];
void gcd(long long a, long long b, long long& d, long long& x, long long& y) {
  if (!b) {
    d = a;
    x = 1;
    y = 0;
    return;
  }
  gcd(b, a % b, d, y, x);
  y -= x * (a / b);
}
long long inv(long long a, long long n) {
  long long d, x, y;
  gcd(a, n, d, x, y);
  return d == 1 ? (x + n) % n : -1;
}
int main() {
  int n, k;
  while (~scanf("%d%d", &n, &k)) {
    k++;
    for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
    val[1] = 1;
    for (int i = 2; i <= n; i++)
      val[i] = val[i - 1] * (k + i - 3) % mod * inv(i - 1, mod) % mod;
    memset(ans, 0, sizeof(ans));
    int t = 0;
    for (int i = 1; i <= n; i++) {
      for (int j = i; j <= n; j++) ans[j] = (ans[j] + a[i] * val[j - t]) % mod;
      t++;
    }
    for (int i = 1; i <= n - 1; i++) cout << ans[i] << " ";
    cout << ans[n] << endl;
  }
}
