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;
vector<long long> v, b;
vector<long long> mul(vector<long long> a, vector<long long> b) {
  vector<long long> r;
  for (int i = 0; i < a.size(); i++) {
    r.push_back(0);
    for (int j = 0; j <= i; j++) {
      r[i] = (r[i] + (a[j] * b[i - j]) % 1000000009LL) % 1000000009LL;
    }
  }
  return r;
}
vector<long long> a;
vector<long long> powMod(vector<long long> a, int k) {
  vector<long long> re = a;
  while (k) {
    if (k & 1) {
      re = mul(re, a);
    }
    a = mul(a, a);
    k = k >> 1;
  }
  return re;
}
int main() {
  long long n, k;
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    long long q;
    cin >> q;
    a.push_back(q);
    b.push_back(1);
  }
  v.push_back(1);
  if (k == 0) {
    for (int i = 0; i < n; i++) cout << a[i] << ' ';
    cout << endl;
    return 0;
  }
  k--;
  v = powMod(b, k);
  vector<long long> ans;
  for (int i = 0; i < n; i++) {
    ans.push_back(0);
    for (int j = 0; j <= i; j++) {
      ans[i] = (ans[i] + (a[j] * v[i - j]) % 1000000009LL) % 1000000009LL;
    }
    cout << ans[i] << ' ';
  }
}
