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> mult(vector<long long> a, vector<long long> b) {
  vector<long long> rez(a.size(), 0);
  for (int i = 0; i < rez.size(); i++) {
    for (int k = 0; k <= i; k++) {
      rez[i] += a[i - k] * b[k];
      rez[i] %= 1000000007;
    }
  }
  return rez;
}
vector<long long> sqr(vector<long long> a) { return mult(a, a); }
vector<long long> binpow(vector<long long> a, long long pow) {
  if (pow == 1) return a;
  if (pow % 2) {
    return mult(sqr(binpow(a, pow / 2)), a);
  } else {
    return sqr(binpow(a, pow / 2));
  }
}
long long n, k;
vector<long long> a;
vector<long long> T;
int main() {
  cin >> n >> k;
  a.resize(n);
  T.resize(n);
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
    T[i] = 1;
  }
  if (k == 0) {
    for (long long i = 0; i < n - 1; i++) cout << a[i] << " ";
    cout << a[n - 1] << endl;
  } else {
    T = binpow(T, k);
    T = mult(T, a);
    for (long long i = 0; i < n - 1; i++) {
      cout << T[i] << " ";
    }
    cout << T[n - 1] << endl;
  }
  return 0;
}
