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;
long long mnoj[3000], sol[2020];
long long n, a[2020], k, A[2020], B[2020], comb[2020];
set<int> prime;
map<int, int> numer, denom;
int gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
void init() {
  cin >> n >> k;
  for (int i = 1; i <= n; i++) cin >> a[i];
  mnoj[1] = 1;
  mnoj[2] = k;
  for (int i = 2; i <= n; i++) {
    A[i] = k + i - 2;
    B[i] = i - 1;
  }
  for (int i = 3; i <= n; i++) {
    for (int j = 2; j <= i; j++) {
      if (B[i] == 1) break;
      int temp =
          gcd(((A[j] > B[i]) ? A[j] : B[i]), ((A[j] < B[i]) ? A[j] : B[i]));
      A[j] /= temp;
      B[i] /= temp;
    }
    mnoj[i] = 1;
    for (int j = 2; j <= i; j++) {
      mnoj[i] = (mnoj[i] * A[j]) % 1000000007LL;
    }
  }
  mnoj[1] = 1;
  sol[1] = a[1];
  for (int i = 2; i <= n; i++) {
    sol[i] = 0;
    for (int j = i, k = 1; j >= 1; j--, k++) {
      sol[i] += (long long)(mnoj[j] * a[k]) % 1000000007LL;
      if (sol[i] >= 1000000007LL) {
        sol[i] %= 1000000007LL;
        if (sol[i] < 0) sol[i] += 1000000007LL;
      }
    }
  }
  for (int i = 1; i <= n; i++) cout << sol[i] << " ";
  cout << endl;
}
int main() {
  init();
  return 0;
}
