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 M = 100010, mod = 1000000007;
long long bigMod(long long first, long long n) {
  if (n == 0) return 1;
  long long temp = bigMod(first, n / 2);
  return n % 2 == 0 ? (temp * temp) % mod
                    : (first * ((temp * temp) % mod)) % mod;
}
long long coef[M];
long long A[M];
int main() {
  ios_base::sync_with_stdio(false);
  long long n, k, i, j;
  cin >> n >> k;
  coef[0] = 1;
  for (i = 1; i <= n; i++) {
    coef[i] = (((coef[i - 1] * k) % mod) * bigMod(i, mod - 2)) % mod;
    k++;
  }
  for (i = 0; i < n; i++) {
    cin >> A[i];
    long long sum = 0;
    for (j = 0, k = i; k >= 0; k--, j++) {
      sum += (coef[j] * A[k]) % mod;
      sum %= mod;
    }
    cout << sum << " ";
  }
  cout << endl;
  return 0;
}
