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 long long mod = 1000000007;
vector<long long> pw(vector<long long> v, int p) {
  if (p == 1) return v;
  vector<long long> res = pw(v, p / 2);
  vector<long long> res2(res.size(), 0);
  for (int i = 0; i < res2.size(); i++)
    for (int j = 0; j <= i; j++)
      res2[i] = (res2[i] + res[j] * res[i - j]) % mod;
  if (p % 2 != 0) {
    for (int i = 1; i < res2.size(); i++)
      res2[i] = (res2[i - 1] + res2[i]) % mod;
  }
  return res2;
}
int main() {
  int n, k, temp;
  cin >> n >> k;
  vector<int> a;
  for (int i = 0; i < n; i++) {
    cin >> temp;
    a.push_back(temp);
  }
  if (k == 0) {
    for (int i = 0; i < n; i++) {
      if (i > 0) cout << " ";
      cout << a[i];
    }
    cout << endl;
    return 0;
  }
  vector<long long> m(n, 1);
  vector<long long> res = pw(m, k);
  for (int i = 0; i < n; i++) {
    long long cur = 0;
    for (int j = 0; j <= i; j++) cur += (a[i - j] * res[j]) % mod;
    cur %= mod;
    if (i > 0) cout << " ";
    cout << cur;
  }
  cout << endl;
  return 0;
}
