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;
long long n, k, a[2005];
long long p[2005];
long long inmod(long long x) {
  long long r = 1;
  for (long long y = 1000000007 - 2;;) {
    if (y & 1) r = (r * x) % 1000000007;
    if (y >>= 1)
      x = (x * x) % 1000000007;
    else
      break;
  }
  return r;
}
int main() {
  scanf("%I64d%I64d", &n, &k);
  long long i, j;
  for (i = 1; i <= n; i++) scanf("%I64d", &a[i]);
  for (i = 1; i <= n; i++) {
    if (k == 0) {
      printf("%I64d ", a[i]);
      continue;
    }
    long long r = 0;
    for (j = i; j >= 1; j--) p[j] = p[j - 1];
    p[1] = (p[2] * (k + i - 2)) % 1000000007;
    long long Q = inmod(i - 1);
    p[1] = (p[1] * Q) % 1000000007;
    if (!p[1] && i == 1) p[1] = 1;
    for (j = 1; j <= i; j++) {
      r = (r + (p[j] * a[j]) % 1000000007) % 1000000007;
    }
    printf("%I64d ", r);
  }
}
