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>
int n, K, lgk, x, y;
long long op[3000][3000], num[3000];
int main() {
  int i, j, k, p, q;
  scanf("%d%d", &n, &K);
  for (i = 1; i <= n; i++) {
    scanf("%I64d", &num[i]);
  }
  lgk = 0;
  for (i = 2; i <= K; i *= 2) {
    lgk++;
  }
  for (i = 1; i <= n; i++) {
    op[0][i] = 1;
  }
  for (i = 1; i <= lgk; i++) {
    for (j = 1; j <= n; j++) {
      op[i][j] = 0;
      for (k = 1; k <= j; k++) {
        op[i][j] += op[i - 1][k] * op[i - 1][j + 1 - k];
        op[i][j] %= 1000000007;
      }
    }
  }
  x = 1;
  for (i = 0; i <= lgk; i++) {
    y = x & K;
    if (y != 0) {
      for (j = n; j >= 1; j--) {
        for (k = 2; k <= j; k++) {
          num[j] += op[i][k] * num[j + 1 - k];
          num[j] %= 1000000007;
        }
      }
    }
    x *= 2;
  }
  for (i = 1; i <= n; i++) {
    printf("%I64d ", num[i]);
  }
  printf("\n");
  getchar();
  getchar();
}
