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>
long long a[2001], b[2001], c[2001];
int n, k;
void mult(long long f1[], long long f2[], long long f3[], int n) {
  for (int i = 0; i < n; i++) {
    f3[i] = 0;
    for (int j = 0; j <= i; j++) {
      f3[i] = (f3[i] + f1[j] * f2[i - j]) % 1000000007;
    }
  }
}
void pow(long long f[], long long o[], int n, int m) {
  int bidx = 0, ridx = 0;
  long long res1[2001] = {0};
  long long res2[2001] = {0};
  long long* res[2] = {res1, res2};
  res[0][0] = 1;
  long long buf1[2001];
  long long buf2[2001];
  long long* buf[2] = {buf1, buf2};
  for (int i = 0; i < n; i++) buf[0][i] = f[i];
  while (m > 0) {
    if (m & 1) {
      mult(res[ridx], buf[bidx], res[(ridx + 1) % 2], n);
      ridx = (ridx + 1) % 2;
    }
    mult(buf[bidx], buf[bidx], buf[(bidx + 1) % 2], n);
    bidx = (bidx + 1) % 2;
    m >>= 1;
  }
  for (int i = 0; i < n; i++) o[i] = res[ridx][i];
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 0; i < n; i++) {
    int tmp;
    scanf("%d", &tmp);
    a[i] = tmp;
  }
  long long ss[2001], rr[2001];
  for (int i = 0; i < n; i++) ss[i] = 1;
  pow(ss, rr, n, k);
  mult(rr, a, ss, n);
  for (int i = 0; i < n; i++) {
    printf("%I64d ", ss[i]);
  }
  printf("\n");
  return 0;
}
