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, a[2001];
unsigned long long s[2001], GT[2001], USED[200001];
const int base = 1e9 + 7;
unsigned long long IndianMul(unsigned long long a, unsigned long long b) {
  if (b == 0) return 0;
  unsigned long long t = IndianMul(a, b / 2);
  if ((b & 1) == 0) t = (t + t);
  if ((b & 1) == 1) t = (t + t + a);
  return t;
}
unsigned long long POW(unsigned long long a, unsigned long long n) {
  if (n == 0) return 1;
  if (n == 1) {
    return a;
  }
  if (n > 1) {
    unsigned long long temp = POW(a, n / 2);
    if ((n & 1) == 0)
      return (temp * temp) % base;
    else
      return ((temp * temp) % base * a) % base;
  }
}
int main() {
  cin >> n >> k;
  for (int i = 1; i <= n; i++) cin >> a[i];
  if (k <= 0)
    for (int i = 1; i <= n; i++) cout << a[i] << " ";
  else if (k > 0) {
    GT[0] = 1;
    for (int i = 1; i <= n; i++) {
      GT[i] = (GT[i - 1] * i) % base;
    }
    USED[0] = 1;
    for (int j = 1; j <= n; j++) {
      for (int i = k; i <= k + n - j - 1; i++) {
        USED[i - k + 1] = (USED[i - k] * i) % base;
      }
    }
    for (int i = 1; i <= n; i++) {
      for (int j = i; j >= 1; j--) {
        unsigned long long x, y;
        x = (USED[i - j] * POW(GT[i - j], base - 2)) % base;
        y = IndianMul(x, a[j]) % base;
        s[i] = (s[i] + y) % base;
      }
      cout << s[i] << " ";
    }
  }
}
