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 mod = 1e9 + 7;
long long n, k, a[2005], c[2005], inv[2005];
long long read() {
  long long f = 0, w = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') w = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    f = f * 10 + ch - '0';
    ch = getchar();
  }
  return f * w;
}
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++) printf("%lld\n", a[i]);
    return 0;
  } else {
    inv[1] = inv[0] = 1;
    for (long long i = 2; i <= n; i++) {
      inv[i] = (mod - mod / i) * inv[mod % i] % mod;
    }
  }
  c[0] = 1;
  for (long long i = 1; i <= n; i++) {
    c[i] = c[i - 1] * (k + i - 1) % mod * inv[i] % mod;
  }
  for (long long i = 1; i <= n; i++) {
    long long ans = 0;
    for (long long j = 1; j <= i; j++) {
      ans = (ans + a[j] * c[i - j] % mod) % mod;
    }
    printf("%lld ", ans);
  }
  return 0;
}
