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 int mxn = 2e3 + 100, mod = 1e9 + 7;
int a[mxn], b[mxn], c[mxn];
int mult(int x, int y) {
  long long ans = 1LL * x * y;
  return ans >= mod ? ans % mod : ans;
}
int add(int x, int y) {
  x += y;
  while (x >= mod) x -= mod;
  return x;
}
int expo(int x, int y) {
  int ret = 1;
  while (y > 0) {
    if (y & 1) ret = mult(ret, x);
    x = mult(x, x);
    y >>= 1;
  }
  return ret;
}
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  for (int i = 0, r = 1, f = 1; i < n; i++) {
    scanf("%d", &a[i]);
    if (i == 0)
      b[i] = 1;
    else {
      r = mult(r, k + i - 1);
      f = mult(f, i);
      b[i] = mult(r, expo(f, mod - 2));
    }
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      if (i + j < n) {
        c[i + j] = add(c[i + j], mult(a[i], b[j]));
      }
    }
  }
  for (int i = 0; i < n; i++) printf("%d ", c[i]);
}
