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>
const int N = 2005, mo = (int)1e9 + 7;
int n, k, a[N], c[N], inv[N];
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  inv[1] = 1;
  for (int i = 2; i <= n; i++) inv[i] = 1LL * (mo - mo / i) * inv[mo % i] % mo;
  for (int i = 0; i < n; i++) {
    c[i] = 1;
    for (int j = 1; j <= i; j++) c[i] = 1LL * c[i] * (k + i - j) % mo;
    for (int j = 1; j <= i; j++) c[i] = 1LL * c[i] * inv[j] % mo;
  }
  for (int i = 1; i <= n; i++) {
    int ans = 0;
    for (int j = 1; j <= i; j++) ans = (ans + 1LL * a[j] * c[i - j]) % mo;
    printf("%d ", ans);
  }
}
