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 int MOD = (int)1e9 + 7;
int a[2100], n, k, ans[2100], d[2100];
int pwd(int a, int n) {
  int ret = 1;
  while (n)
    if (n & 1)
      n--, ret = (long long)ret * a % MOD;
    else
      n >>= 1, a = (long long)a * a % MOD;
  return ret;
}
int inv(int a) { return pwd(a, MOD - 2); }
int main() {
  cin >> n >> k;
  for (int i = (1); i <= (n); i++) scanf("%d", a + i);
  for (int i = (1); i <= (n); i++) {
    d[i] = 1;
    for (int j = (k); j <= (k + i - 2); j++) d[i] = (long long)d[i] * j % MOD;
    for (int j = (1); j <= (i - 1); j++) d[i] = (long long)d[i] * inv(j) % MOD;
  }
  for (int i = (1); i <= (n); i++)
    for (int j = (i); j <= (n); j++)
      ans[j] = (ans[j] + (long long)d[j - i + 1] * a[i]) % MOD;
  for (int i = (1); i <= (n); i++) printf("%d ", ans[i]);
  puts("");
  return 0;
}
