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;
inline long long read() {
  long long x = 0;
  char ch = getchar();
  bool f = false;
  while (!isdigit(ch)) f |= ch == '-', ch = getchar();
  while (isdigit(ch)) x = (x << 1) + (x << 3) + ch - 48, ch = getchar();
  return f ? -x : x;
}
void write(long long x) {
  if (x < 0) x = -x, putchar('-');
  if (x > 9) write(x / 10);
  putchar(x % 10 + 48);
}
const long long dev = 1e9 + 7;
long long n, x, y, A[2005], C[2005];
long long QP(long long un, long long up) {
  long long res = 1;
  for (; up; up >>= 1, un = un * un % dev)
    if (up & 1) res = res * un % dev;
  return res;
}
int main() {
  n = read();
  x = y = read() - 1;
  C[0] = 1;
  for (int i = 1; i <= n; i++) A[i] = read();
  for (int i = 1; i <= n; i++)
    x++, C[i] = C[i - 1] * x % dev * QP(x - y, dev - 2) % dev;
  for (int i = 1; i <= n; i++) {
    long long res = 0;
    for (int j = 0; j < i; j++) res = (res + A[i - j] * C[j] % dev) % dev;
    (write(res), putchar(' '));
  }
  return 0;
}
