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>
long long mod = 1000ll * 1000ll * 1000ll + 7;
long long p[2000], inv[2050], f[2050], ans, i, j, k, n;
long long C(long long n, long long m) {
  long long i;
  long long ans = 1;
  for (i = 0; i < m; i++) ans = (ans * (inv[i + 1] * (n - i) % mod)) % mod;
  return ans;
}
int main() {
  inv[0] = 1;
  inv[1] = 1;
  for (int i = 2; i < 2050; i++) {
    inv[i] = (-mod / i) * inv[mod % i];
    inv[i] = (inv[i] % mod + mod) % mod;
  }
  scanf("%I64d%I64d", &n, &k);
  for (i = 0; i < n; i++) scanf("%I64d", &f[i]);
  if (k == 0) {
    for (i = 0; i < n; i++) printf("%I64d ", f[i]);
    puts("");
    return 0;
  }
  k--;
  for (i = 0; i < n; i++) {
    p[i] = C(i + k, i);
  }
  for (i = 0; i < n; i++) {
    ans = 0;
    for (j = 0; j <= i; j++) ans = (ans + p[i - j] * f[j] % mod) % mod;
    printf("%I64d ", ans);
  }
  puts("");
  return 0;
}
