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;
long double pi = 3.141592653589793238;
const long long INF = long long(2e18);
long long mo(const long long input, const long long ceil) {
  return input >= ceil ? input % ceil : input;
}
void solve(int tc) {
  long long n, k;
  cin >> n >> k;
  vector<long long> a(n + 5, 0);
  for (long long i = 1; i <= n; i++) {
    cin >> a[i];
  }
  vector<long long> cum(n + 5, 0);
  for (long long i = 1; i <= n; i++) {
    cum[i] = (cum[i - 1] + a[i]) % 1000000007;
  }
  for (long long i = 1; i <= n; i++) {
    long long df = a[i];
    long long ad = (k * cum[i - 1]) % 1000000007;
    df += ad;
    df %= 1000000007;
    cout << df << " ";
  }
  cout << "\n";
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int tt = 1;
  for (int i = 0; i < tt; i++) {
    solve(i + 1);
  }
  return 0;
}
