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>
const int N = 2005;
const int Mod = 1000000007;
long long a[N], x[N], iv[N];
void ee(long long a, long long b, long long &d, long long &x, long long &y) {
  if (!b) {
    d = a, x = 1, y = 0;
  } else {
    ee(b, a % b, d, y, x);
    y -= x * (a / b);
  }
}
long long inv(long long a, long long n) {
  long long d, x, y;
  ee(a, n, d, x, y);
  return (d == 1 ? (x + n) % n : -1);
}
int main() {
  int n, k;
  long long sum;
  for (int i = 1; i < N; i++) iv[i] = inv(i, Mod);
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%I64d", &a[i]);
  x[0] = 1;
  for (int i = 1; i < n; i++) x[i] = x[i - 1] * (k - 1 + i) % Mod * iv[i] % Mod;
  for (int i = 1; i <= n; i++) {
    sum = 0;
    for (int j = 1; j <= i; j++) sum = (sum + x[i - j] * a[j] % Mod) % Mod;
    printf("%I64d ", sum);
  }
  return 0;
}
