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 maxn = 2035;
int n, k, a[maxn], b[maxn], inv[maxn], fac[maxn], pre[maxn];
void init() {
  fac[0] = fac[1] = inv[0] = inv[1] = 1;
  for (int i = 2; i <= n; i++)
    fac[i] = 1ll * fac[i - 1] * i % 1000000007,
    inv[i] =
        1000000007 - 1ll * (1000000007 / i) * inv[1000000007 % i] % 1000000007;
  pre[0] = 1;
  for (int i = 1; i <= n; i++)
    pre[i] = 1ll * pre[i - 1] * (k % 1000000007 + i - 1) % 1000000007 * inv[i] %
             1000000007;
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]), b[i] = a[i];
  if (k) {
    init();
    for (int i = 1; i <= n; i++) {
      b[i] = 0;
      for (int j = 1; j <= i; j++)
        b[i] = 1ll * (b[i] + 1ll * pre[i - j] * a[j] % 1000000007) % 1000000007;
    }
  }
  for (int i = 1; i <= n; i++) printf("%d ", b[i]);
  return 0;
}
