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;
int n, k;
const int Mod = 1e9 + 7;
int a[2010];
long long init[2010];
long long now[2020], d[2020];
void calc() {
  for (int i = 1; i <= n; i++) {
    d[i] = now[i];
    now[i] = 0;
  }
  for (int i = 1; i <= n; i++) {
    for (int j = i; j >= 1; j--) {
      now[i] = (now[i] + d[i - j + 1] * init[j]) % Mod;
    }
  }
}
void mo() {
  for (int i = 1; i <= n; i++) {
    d[i] = init[i];
    init[i] = 0;
  }
  for (int i = 1; i <= n; i++) {
    for (int j = i; j >= 1; j--) {
      init[i] = (init[i] + d[i - j + 1] * d[j]) % Mod;
    }
  }
}
void deal(long long k) {
  k--;
  while (k) {
    if (k & 1) calc();
    k = k / 2;
    mo();
  }
}
int main() {
  cin >> n >> k;
  for (int i = 1; i <= n; i++) cin >> a[i];
  if (k == 0) {
    for (int i = 1; i <= n; i++) cout << a[i] << " ";
    cout << endl;
  } else {
    for (int i = 1; i <= n; i++) init[i] = now[i] = 1;
    deal(k);
    for (int i = 1; i <= n; i++) {
      long long ans = 0;
      for (int j = 1; j <= i; j++) {
        ans = (ans + a[j] * now[i - j + 1]) % Mod;
      }
      cout << ans << " ";
    }
  }
  return 0;
}
