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 long long infl = 1e18 + 5;
long long int n, m, k, q, x, y, f, val, t = 1, i, j;
long long int ind = -1, cnt, sz, sm, ans, mx = -1, mn = infl;
long long int a[2004], rfact[2004], fact[2004], b[2004];
long long int modpow(long long int a, long long int b) {
  long long int res = 1;
  while (b > 0) {
    if (b & 1) {
      res *= a;
      res %= 1000000007;
    }
    b /= 2LL;
    a = a * a;
    a %= 1000000007;
  }
  return res;
}
long long int nci(long long int n, long long int r) {
  if (n < r || r > n) return 0;
  n -= k;
  long long int den = modpow(rfact[r], 1000000007 - 2);
  long long int val = fact[n] * den;
  val %= 1000000007;
  return val;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  if (fopen("inp.txt", "r")) {
    freopen("myfile.txt", "w", stdout);
    freopen("inp.txt", "r", stdin);
  }
  cin >> n >> k;
  for (i = 0; i < n; i++) {
    cin >> a[i];
  }
  fact[0] = k;
  for (i = 1; i < 2004; i++) {
    fact[i] = (k + i) * fact[i - 1];
    fact[i] %= 1000000007;
  }
  rfact[0] = 1;
  for (i = 1; i < 2004; i++) {
    rfact[i] = i * rfact[i - 1];
    rfact[i] %= 1000000007;
  }
  b[0] = a[0];
  for (i = 1; i < n; i++) {
    b[i] = a[i] + (k * a[i - 1]) % 1000000007;
    b[i] %= 1000000007;
    for (j = i - 2; j >= 0; j--) {
      long long int temp = nci(k + i - j - 1, i - j);
      val = temp * a[j];
      val %= 1000000007;
      b[i] += val;
      b[i] %= 1000000007;
    }
  }
  for (i = 0; i < n; i++) cout << b[i] << " ";
  return 0;
}
