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 NN = 2000 + 10;
const long long Mod = 1000000007;
int N, K;
long long a[NN];
long long eye[NN], init[NN];
void doit(long long *eye, long long *init) {
  long long mid[NN];
  for (int i = 0; i < N; i++) {
    mid[i] = 0;
    for (int j = 0; j <= i; j++) {
      mid[i] += eye[i - j] * init[j] % Mod;
      if (mid[i] >= Mod) mid[i] -= Mod;
    }
  }
  for (int i = 0; i < N; i++) eye[i] = mid[i];
}
void calc(int n) {
  for (int i = 0; i < N; i++) eye[i] = 0, init[i] = 1;
  eye[0] = 1;
  while (n) {
    if (n & 1) doit(eye, init);
    n >>= 1;
    doit(init, init);
  }
}
int main() {
  while (scanf("%d %d", &N, &K) == 2) {
    for (int i = 0; i < N; i++) scanf("%lld", &a[i]);
    calc(K);
    for (int i = 0; i < N; i++) {
      long long ans = 0;
      for (int j = 0; j <= i; j++) {
        ans += eye[i - j] * a[j];
        if (ans >= Mod) ans -= Mod;
      }
      if (i != 0) printf(" ");
      printf("%lld", ans);
    }
    printf("\n");
  }
  return 0;
}
