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;
vector<int> prod(const vector<int> a, const vector<int> b) {
  vector<int> ans(n);
  for (int i = 0; i < n; ++i)
    for (int j = 0; j <= i; ++j)
      ans[i] = (ans[i] + 1LL * a[j] * b[i - j]) % 1000000007;
  return ans;
}
vector<int> pown(const vector<int> a, int k) {
  if (k == 0) {
    vector<int> one(n);
    one[0] = 1;
    return one;
  }
  if (k == 1) return a;
  vector<int> ans = pown(a, k / 2);
  ans = prod(ans, ans);
  if (k % 2) ans = prod(ans, a);
  return ans;
}
bool solve() {
  if (scanf("%d%d", &n, &k) == EOF) return false;
  vector<int> a(n);
  for (int i = 0; i < n; ++i) scanf("%d", &a[i]);
  vector<int> ans = prod(a, pown(vector<int>(n, 1), k));
  for (int i = 0; i < n; ++i) printf("%d ", ans[i]);
  printf("\n");
  return true;
}
int main() {
  solve();
  return 0;
}
