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>
int a[2000];
long long B[2000];
void extended_gcd(long long a, long long b, long long &i, long long &j) {
  if (b == 0) {
    i = 1;
    j = 0;
    return;
  }
  extended_gcd(b, a % b, j, i);
  j = ((j - (a / b) * i) % 1000000007 + 1000000007) % 1000000007;
}
void work(int n, int k) {
  long long a = 1, b = 1, I, J;
  B[0] = 1;
  for (int i = 1; i < n; i++) {
    a = (a * (k + i - 1)) % 1000000007;
    b = (b * i) % 1000000007;
    extended_gcd(b, 1000000007, I, J);
    B[i] = (a * I) % 1000000007;
  }
}
int main() {
  int n, k;
  scanf("%d %d", &n, &k);
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  work(n, k);
  for (int i = 0; i < n; i++) {
    int tmp = 0;
    for (int j = 0; j <= i; j++) {
      tmp += (B[i - j] * a[j]) % 1000000007;
      if (tmp >= 1000000007) tmp -= 1000000007;
    }
    printf((i) ? " %d" : "%d", tmp);
  }
  puts("");
}
