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 long power(long long x, long long n, long long M) {
  long long result = 1;
  while (n > 0) {
    if (n % 2 == 1) result = (result * x) % M;
    x = (x * x) % M;
    n = n / 2;
  }
  return result;
}
void solve() {
  long long n, k;
  cin >> n >> k;
  vector<long long> a(n);
  for (long long i = 0; i < n; i++) cin >> a[i];
  if (k == 0) {
    for (long long i = 0; i < n; i++) cout << a[i] << " ";
    return;
  }
  cout << a[0] << " ";
  vector<long long> foo(2000);
  k--;
  for (long long i = 0; i < 2000; i++) {
    long long nume = 1, cnt = i;
    for (long long j = k; j >= 1; j--) {
      if (cnt <= 0) break;
      nume = (nume * j) % 1000000007;
      cnt--;
    }
    long long deno = 1;
    for (long long j = i; j >= 1; j--) {
      deno = (deno * j) % 1000000007;
    }
    foo[i] = (nume * power(deno, 1000000007 - 2, 1000000007)) % 1000000007;
    k++;
  }
  for (long long i = 1; i < n; i++) {
    long long ans = 0, cnt = 0;
    for (long long j = i; j >= 0; j--) {
      ans = (ans + (a[j] * foo[cnt]) % 1000000007) % 1000000007;
      cnt++;
    }
    cout << ans << " ";
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long t;
  t = 1;
  while (t--) solve();
}
