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;
int c[2000];
long long inv[3000];
long long ter[2000];
long long ret[2000];
int main() {
  int mod = 1000000007;
  inv[1] = 1LL;
  for (int i = 2; i < 3000; i++) {
    inv[i] = mod - inv[mod % i] * (mod / i) % mod;
  }
  int a, b;
  scanf("%d%d", &a, &b);
  for (int i = 0; i < a; i++) scanf("%d", c + i);
  if (b == 0) {
    for (int i = 0; i < a; i++) {
      printf("%d", c[i]);
      if (i < a - 1)
        printf(" ");
      else
        printf("\n");
    }
    return 0;
  }
  b--;
  for (int i = 0; i < a; i++) {
    ter[i] = 1LL;
    for (int j = 0; j < i; j++) {
      ter[i] = ter[i] * (b + i - j) % mod;
      ter[i] = ter[i] * inv[j + 1] % mod;
    }
  }
  for (int i = 0; i < a; i++) {
    for (int j = 0; i + j < a; j++) {
      ret[i + j] = (ret[i + j] + ter[j] * c[i]) % mod;
    }
  }
  for (int i = 0; i < a; i++) {
    printf("%d", (int)(ret[i]));
    if (i < a - 1)
      printf(" ");
    else
      printf("\n");
  }
}
