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 long long MAXN = 2010;
const long long mod = 1e9 + 7;
long long n, k, a[MAXN], c[MAXN], inv[MAXN];
inline long long read() {
  long long X = 0, flag = 0;
  char ch = 0;
  while (!isdigit(ch)) flag |= ch == '-', ch = getchar();
  while (isdigit(ch)) X = (X << 3) + (X << 1) + (ch ^ 48), ch = getchar();
  return flag ? -X : X;
}
inline void write(long long X) {
  if (X < 0) X = -X, putchar('-');
  if (X > 9) write(X / 10);
  putchar(X % 10 + '0');
}
signed main() {
  n = read(), k = read();
  for (long long i = 1; i <= n; ++i) a[i] = read();
  if (!k) {
    for (long long i = 1; i <= n; ++i) write(a[i]), putchar(' ');
    return 0;
  }
  c[0] = 1, inv[0] = inv[1] = 1;
  for (long long i = 2; i <= n; ++i)
    inv[i] = inv[mod % i] * (mod - mod / i) % mod;
  for (long long i = 1; i <= n; ++i)
    c[i] = c[i - 1] * inv[i] % mod * (i + k - 1) % mod;
  for (long long i = 1, num; i <= n; ++i) {
    num = 0;
    for (long long j = 1; j <= i; ++j) (num += a[j] * c[i - j] % mod) %= mod;
    write(num), putchar(' ');
  }
  return 0;
}
