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 M = 1000000007LL;
const int maxn = 2000 + 20;
long long c[maxn];
int a[maxn];
long long qm(long long x, long long y) {
  long long t = 1;
  while (y > 0) {
    while (y % 2 == 0) {
      x = x * x % M;
      y /= 2;
    }
    t = t * x % M;
    y--;
  }
  return t;
}
int main() {
  int n, k;
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    if (k == 0) cout << a[i] << " ";
  }
  if (k == 0) exit(0);
  memset(c, 0, sizeof(c));
  c[1] = 1;
  for (int i = 2; i <= n; i++)
    c[i] = (c[i - 1] * (k + i - 2)) % M * qm(i - 1, M - 2) % M;
  for (int i = 1; i <= n; i++) {
    long long ans = 0;
    for (int j = 1; j <= i; j++) ans = (ans + c[j] * a[i - j + 1]) % M;
    cout << ans << " ";
  }
  return 0;
}
