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>
#pragma comment(linker, "/STACK:16777216")
using namespace std;
const int MAX_N = 2004;
const long long MODUN = 1000000007LL;
long long v[MAX_N], f[MAX_N];
int read() {
  int x;
  scanf("%d", &x);
  return x;
}
long long mmod(long long x) {
  x %= MODUN;
  return (x + MODUN) % MODUN;
}
long long mpow(long long x, long long a) {
  if (a == 0) return 1;
  if (a == 1) return mmod(x);
  long long t = mpow(x, a / 2);
  t = mmod(t * t);
  if (a % 2 == 1) return mmod(t * x);
  return t;
}
int main() {
  long long n, k;
  cin >> n >> k;
  --k;
  for (int i = 0; i < int(n); ++i) scanf("%I64d", &v[i]);
  memset(f, (0), sizeof(f));
  f[0] = 1;
  long long a = 1, b = 1;
  for (long long i = 1; i <= n; ++i) {
    a = mmod(a * i);
    b = mmod(b * (k + i));
    f[i] = mmod(b * mpow(a, MODUN - 2));
  }
  for (int i = 0; i < int(n); ++i) {
    long long s = 0;
    for (int j = int(i); j >= int(0); --j) s = mmod(s + mmod(f[i - j] * v[j]));
    printf("%I64d ", s);
  }
  return 0;
}
