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;
long long a = 1, b = 1, x, y, pcnt = 0;
long long extended_gcd(long long a, long long b) {
  if (!b) {
    x = 1, y = 0;
    return a;
  }
  long long ret = extended_gcd(b, a % b), tmp = x;
  x = y, y = tmp - (a / b) * y;
  return ret;
}
unsigned long long k, n, v[2002], c[2002], s[2002];
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 0; i < n; ++i) cin >> v[i];
  if (k) {
    k--;
    c[0] = 1;
    for (int i = 1; i < n; ++i) {
      a *= k + i, b *= i;
      while (a % 1000000007 == 0) a /= 1000000007, pcnt++;
      while (b % 1000000007 == 0) b /= 1000000007, pcnt--;
      b %= 1000000007, a %= 1000000007;
      if (pcnt)
        c[i] = 0;
      else {
        extended_gcd(b, 1000000007);
        if (x < 0) x += 1000000007;
        x *= a, x %= 1000000007;
        c[i] = x;
      }
    }
    for (int i = 0; i < n; ++i)
      for (int j = 0; j < i + 1; ++j)
        s[i] = (s[i] + v[i - j] * c[j]) % 1000000007;
    for (int i = 0; i < n; ++i) cout << s[i] << " ";
  } else
    for (int i = 0; i < n; ++i) cout << v[i] << " ";
  return 0;
}
