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;
template <class T>
bool umin(T& a, T b) {
  if (a > b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
bool umax(T& a, T b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
long long arr[100009], comb[100009];
long long mod(long long x) { return (x % 1000000007); }
long long Fe(int x, int y) {
  if (!y) return 1LL;
  long long h = Fe(x, y / 2);
  h = mod(h * h);
  if (y & 1) h = mod(h * x);
  return h;
}
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%lld", arr + i);
  if (k == 0) {
    for (int i = 1; i <= n; i++) printf("%lld ", arr[i]);
    printf("\n");
    return 0;
  }
  if (k == 1) {
    for (int i = 1; i <= n; i++) {
      arr[i] = mod(arr[i] + arr[i - 1]);
      printf("%lld ", arr[i]);
    }
    printf("\n");
    return 0;
  }
  long long C = 1, cnt = 1;
  for (int i = k - 1; i <= (k - 1) + n; i++) {
    comb[cnt] = C;
    C = mod(C * (i + 1));
    C = mod(C * Fe(cnt, 1000000007 - 2));
    cnt++;
  }
  for (int i = 1; i <= n; i++) {
    long long ans = 0;
    for (int j = i; j >= 1; j--) ans = mod(ans + mod(arr[i - j + 1] * comb[j]));
    printf("%lld ", ans);
  }
  return 0;
}
