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 int N = 2010;
const int B = 1e9 + 7;
int n, k;
long long a[N], b[N];
long long tmp[N];
long long h[N];
void mul(long long *a, long long *b) {
  memset(tmp, 0, sizeof(tmp));
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= i; ++j) {
      tmp[i] = (tmp[i] + a[j] * b[i - j + 1]) % B;
    }
  }
  for (int i = 1; i <= n; ++i) a[i] = tmp[i];
}
void getmatrix() {
  for (int i = 1; i <= n; ++i) h[i] = 1;
  memset(b, 0, sizeof(b));
  b[1] = 1;
  int tk = k;
  while (tk) {
    if (tk & 1) mul(b, h);
    mul(h, h);
    tk >>= 1;
  }
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; ++i) {
    scanf("%I64d", a + i);
  }
  getmatrix();
  for (int i = 1; i <= n; ++i) {
    long long x = 0;
    for (int j = 1; j <= i; ++j) {
      x = (x + a[i - j + 1] * b[j]) % B;
    }
    printf("%I64d%c", x, " \n"[i == n]);
  }
  return 0;
}
