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;
template <class T>
inline void read(T& x) {
  bool fu = 0;
  char c;
  for (c = getchar(); c <= 32; c = getchar())
    ;
  if (c == '-') fu = 1, c = getchar();
  for (x = 0; c > 32; c = getchar()) x = x * 10 + c - '0';
  if (fu) x = -x;
};
template <class T>
inline void read(T& x, T& y) {
  read(x);
  read(y);
}
template <class T>
inline void read(T& x, T& y, T& z) {
  read(x);
  read(y);
  read(z);
}
inline char getc() {
  char c;
  for (c = getchar(); c <= 32; c = getchar())
    ;
  return c;
}
const int mod = int(1e9) + 7;
int n, K, i, j, k, l, p;
long long a[3010], s[3010], fac[3010], ic[3010], z[3010];
long long pow(long long a, long long b, long long c) {
  long long ans = 1;
  for (a %= c; b; b >>= 1, a = a * a % c)
    if (b & 1) ans = ans * a % c;
  return ans;
}
int main() {
  read(n, K);
  for (i = 0; i <= 2000; i++) ic[i] = pow(i, mod - 2, mod);
  for (j = 0; j <= 2000; j++) {
    long long res = 1;
    for (i = 1; i <= j; i++) res = res * (K + j - i) % mod * ic[i] % mod;
    z[j] = res;
  }
  for (i = 1; i <= n; i++) read(a[i]);
  for (i = 1; i <= n; i++) {
    for (j = i; j <= n; j++) s[j] = (s[j] + z[j - i] * a[i]) % mod;
  }
  for (i = 1; i <= n; i++) printf("%I64d ", s[i]);
  printf("\n");
  scanf("\n");
  return 0;
}
