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 long long MOD = 1000000007;
long long a[N], tmp[N], ret[N], buf[N];
long long n, k;
void mul(long long *a, const long long *b) {
  for (int i = 0; i < n; i++) buf[i] = 0;
  for (int i = 0; i < n; i++)
    for (int j = i; j >= 0; j--) buf[i] += a[i - j] * b[j], buf[i] %= MOD;
  for (int i = 0; i < n; i++) a[i] = buf[i];
}
void add(long long *a, const long long *b) {
  for (int i = 0; i < n; i++) a[i] += b[i];
}
int main() {
  cin >> n >> k;
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < n; i++) tmp[i] = 1, ret[i] = 0;
  ret[0] = 1;
  for (long long i = 1; i <= k; i <<= 1) {
    if (k & i) {
      mul(ret, tmp);
    }
    mul(tmp, tmp);
  }
  mul(a, ret);
  for (int i = 0; i < n; i++) {
    if (i) putchar(' ');
    cout << a[i];
  }
  return 0;
}
