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;
const long long mod = 1000000007;
const int maxn = 2015;
long long ck[maxn], v[maxn], ans[maxn];
long long pow_mod(long long v, long long st) {
  long long a = v, ans = 1;
  for (; st; st /= 2) {
    if (st & 1) ans = (ans * a) % mod;
    a = (a * a) % mod;
  }
  return ans;
}
long long get_obr(long long v) { return pow_mod(v, mod - 2); }
int main() {
  long long i, j, k, n, tmp;
  cin >> n >> k;
  for (i = 1; i <= n; i++) cin >> v[i];
  ck[0] = 1;
  for (i = 1; i < n; i++) {
    ck[i] = (ck[i - 1] * (k - 1 + i)) % mod;
    tmp = get_obr(i);
    ck[i] = (ck[i] * tmp) % mod;
  }
  for (i = 1; i <= n; i++) {
    for (j = 0; j <= i; j++) ans[i] = (ans[i] + v[j + 1] * ck[i - 1 - j]) % mod;
  }
  for (i = 1; i <= n; i++) cout << ans[i] << " ";
  return 0;
}
