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;
const long long d = 1000000007;
int n, k, a[2222];
long long pd(long long a, long long b) {
  long long c = 1;
  while (b) {
    if (b & 1) c = (c * a) % d;
    a = (a * a) % d;
    b >>= 1;
  }
  return c;
}
long long fc[2222], ifc[2222], ic[2222];
int b[2222];
int main() {
  fc[0] = 1;
  ifc[0] = 1;
  ic[0] = 0;
  for (int i = (1); i <= (2100); ++i) {
    fc[i] = fc[i - 1] * i % d;
    ifc[i] = pd(fc[i], d - 2);
    ic[i] = pd(i, d - 2);
  }
  scanf("%d%d", &n, &k);
  for (int i = (1); i <= (n); ++i) scanf("%d", a + i);
  if (!k) {
    for (int i = (1); i <= (n); ++i) printf("%d%s", a[i], i == n ? "\n" : " ");
    return 0;
  }
  --k;
  for (int p = (1); p <= (n); ++p) {
    int ans = 0;
    long long t = 1;
    for (int i = (p); i >= (1); --i) {
      (ans += a[i] * t) %= d;
      (t *= ic[p - i + 1]) %= d;
      (t *= p - (i - 1) + k) %= d;
    }
    printf("%d%s", ans, p == n ? "\n" : " ");
  }
  return 0;
}
