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;
int const mod = 1e9 + 7;
int const maxn = 2005;
inline void safe_mul(int& a, int b) { a = (a * 1LL * b) % mod; }
inline int mypow(int a, int b) {
  int ans = 1;
  int tmp = a;
  while (b) {
    if (b & 1) safe_mul(ans, tmp);
    safe_mul(tmp, tmp);
    b >>= 1;
  }
  return ans;
}
inline int inv(int x) { return mypow(x, mod - 2); }
int koef[maxn];
int a[maxn];
int main() {
  ios_base::sync_with_stdio(false);
  int n, k;
  cin >> n >> k;
  koef[0] = 1;
  for (int i = 1; i < n; ++i) {
    koef[i] = koef[i - 1];
    safe_mul(koef[i], (k + i - 1) * 1LL * inv(i) % mod);
  }
  for (int i = 0; i < n; ++i) cin >> a[i];
  for (int i = 0; i < n; ++i) {
    int ans = 0;
    for (int j = 0; j <= i; ++j) ans = (ans + koef[j] * 1LL * a[i - j]) % mod;
    cout << ans << " \n"[i == n - 1];
  }
  return 0;
}
