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 n, k;
int data[2005];
static inline int mult(long long int a, long long int b) {
  long long res = (a * b) % 1000000007;
  return res;
}
static inline int inverse(long long a) {
  long long ans = 1;
  while (a > 1) {
    long long k = 1000000007 / a + 1;
    a = k * a - 1000000007;
    ans = mult(ans, k);
  }
  return ans;
}
long long koef[2004];
int ans[2004];
int main() {
  for (int i = 1; i <= 2001; i++) {
  }
  scanf("%d %d", &n, &k);
  for (int i = 0; i < n; i++) {
    scanf("%d", &data[i]);
  }
  if (k > 0) {
    koef[0] = 1;
    for (int i = 1; i < n; i++) {
      koef[i] = mult(mult(koef[i - 1], k + i - 1), inverse(i));
    }
    for (int i = 0; i < n; i++) {
    }
    for (int i = 0; i < n; i++) {
      ans[i] = data[i];
      for (int j = 1; j <= i; j++) {
        ans[i] = (ans[i] + mult(data[i - j], koef[j])) % 1000000007;
      }
    }
  } else {
    memcpy(ans, data, sizeof(ans));
  }
  printf("%d", ans[0]);
  for (int i = 1; i < n; i++) printf(" %d", ans[i]);
  printf("\n");
  return 0;
}
