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;
namespace p224EPartialSums {
const int k_mod = 1000000007;
long long arr[2010] = {0}, sum[2010] = {0};
long long pt[2010], temp[2010];
int inv[2010];
inline int absmod(int val) {
  val %= k_mod;
  return val < 0 ? val + k_mod : val;
}
void solve(int test_num) {
  int len, reps;
  scanf("%d %d", &len, &reps);
  for (int i = 0; i < len; i++) {
    scanf("%d", &arr[i]);
    temp[i] = arr[i];
  }
  for (int i = 1; i <= len; i++) {
    int a = i, b = k_mod;
    int x = 0, y = 1, u = 1, v = 0;
    while (a != 0) {
      int q = b / a, r = b % a;
      int m = x - u * q, n = y - v * q;
      b = a;
      a = r;
      x = u;
      y = b;
      u = m;
      v = n;
    }
    assert(b == 1);
    inv[i] = absmod(x);
  }
  pt[0] = 1;
  long long num = reps, den = 1;
  for (int i = 1; i <= len; i++) {
    pt[i] = ((pt[i - 1] * num) % k_mod * inv[den]) % k_mod;
    num++;
    den++;
  }
  for (int i = 0; i < len; i++) {
    for (int j = 0; j <= i; j++) {
      sum[i] += (arr[j] * pt[i - j]) % k_mod;
      if (sum[i] >= k_mod) sum[i] -= k_mod;
    }
  }
  for (int i = 0; i < len; i++) printf("%d ", sum[i]);
}
void solve() { solve(1); }
}  // namespace p224EPartialSums
int main() {
  p224EPartialSums::solve();
  return 0;
}
