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;
#pragma comment(linker, "/STACK:128000000")
const long long mod = 1e9 + 7;
const int MAXN = 2010;
int n, k;
vector<long long> a;
long long t;
vector<bool> bits;
vector<long long> v, v1;
inline void init() {
  for (int i = 0; i < n; i++) {
    v.push_back(0);
    v1.push_back(1);
  }
  v[0] = 1;
}
inline vector<long long> multiply(vector<long long> &a, vector<long long> &b) {
  vector<long long> ans;
  ans.assign(n, 0);
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      if (i + j >= n) break;
      ans[i + j] += a[i] * b[j];
      ans[i + j] %= mod;
    }
  }
  return ans;
}
int main() {
  cin >> n >> k;
  int kt = k;
  while (kt) {
    if (kt & 1)
      bits.push_back(true);
    else
      bits.push_back(false);
    kt >>= 1;
  }
  init();
  while (!bits.empty()) {
    v = multiply(v, v);
    if (bits.back()) v = multiply(v, v1);
    bits.pop_back();
  }
  for (int i = 0; i < n; i++) {
    cin >> t;
    a.push_back(t);
  }
  a = multiply(a, v);
  for (int i = 0; i < n; i++) cout << a[i] << ' ';
  return 0;
}
