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 maxN = 2000 + 10, Mod = 1000 * 1000 * 1000 + 7;
int n, k, arr[maxN], ent[maxN];
int pw(int a, int b) {
  long long ans = 1;
  for (int i = 31; i >= 0; i--) {
    ans *= ans;
    ans %= Mod;
    if ((b >> i) & 1) ans *= a;
    ans %= Mod;
  }
  return ans;
}
int C(int a, int r) {
  long long rf = 1, af = 1;
  for (int i = 2; i <= r; i++) {
    rf *= i;
    rf %= Mod;
  }
  for (int j = a; j > a - r; j--) {
    af *= j;
    af %= Mod;
  }
  long long ans = af * (long long)pw(rf, Mod - 2);
  return ans % Mod;
}
int main() {
  cin >> n >> k;
  for (int i = 0; i < n; i++) cin >> arr[i];
  if (k == 0) {
    for (int i = 0; i < n; i++) cout << arr[i] << ' ';
    cout << endl;
    return 0;
  }
  k--;
  for (int i = 0; i < n; i++) ent[i] = C(k + i, i);
  for (int i = 0; i < n; i++) {
    long long ans = 0;
    for (int j = 0; j <= i; j++) {
      ans += (long long)ent[i - j] * (long long)arr[j];
      ans %= Mod;
    }
    cout << ans << ' ';
  }
  cout << endl;
  return 0;
}
