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;
long long int t[2001];
long long int c[2001] = {0};
void multiply(long long int a[], long long int b[], long long int n) {
  memset(c, 0, sizeof(c));
  for (int i = 0; i < n; i++) {
    for (int j = 0; i < n; j++) {
      if (j > i) break;
      c[i] += a[i - j] * b[j];
      c[i] %= 1000000007;
    }
  }
  for (int i = 0; i < n; i++) a[i] = c[i];
}
void rec(long long int b[], long long int k, long long int n) {
  if (k <= 1) return;
  rec(b, k / 2, n);
  multiply(b, b, n);
  if (k & 1) multiply(b, t, n);
}
int main() {
  long long int n, k;
  cin >> n >> k;
  long long int a[n];
  for (int i = 0; i < n; i++) cin >> a[i];
  long long int b[n];
  for (int i = 0; i < n; i++) {
    b[i] = 1;
    t[i] = 1;
  }
  if (k == 0) {
    for (int i = 0; i < n; i++) cout << a[i] % 1000000007 << " ";
    return 0;
  }
  rec(b, k, n);
  multiply(b, a, n);
  for (int i = 0; i < n; i++) cout << b[i] % 1000000007 << " ";
}
