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;
long long Ex_Gcd(long long a, long long b, long long &x, long long &y) {
  long long r, t;
  if (b == 0) {
    x = 1;
    y = 0;
    return a;
  }
  r = Ex_Gcd(b, a % b, x, y);
  t = x;
  x = y;
  y = t - a / b * y;
  return r;
}
long long inv[2010];
long long x, y;
long long n, k;
long long c[2010];
long long ans[2010];
int main() {
  for (int i = 2; i <= 2005; i++) {
    Ex_Gcd(i, 1000000007, x, y);
    x = (x % 1000000007 + 1000000007) % 1000000007;
    inv[i] = x;
  }
  inv[1] = 1;
  scanf("%I64d%I64d", &n, &k);
  for (int i = 1; i <= n; i++) ans[i] = 0;
  c[1] = 1;
  if (k == 0) {
    for (int i = 1; i <= n; i++) scanf("%I64d", &ans[i]);
    printf("%I64d", ans[1]);
    for (int i = 2; i <= n; i++) printf(" %I64d", ans[i]);
    return 0;
  }
  for (int i = 2; i <= n; i++) {
    c[i] = (((c[i - 1] * inv[i - 1]) % 1000000007) * k) % 1000000007;
    k++;
  }
  for (int i = 1; i <= n; i++) {
    scanf("%I64d", &x);
    ans[i] = (ans[i] + x) % 1000000007;
    for (int j = 1; i + j <= n; j++)
      ans[i + j] = (ans[i + j] + c[j + 1] * x % 1000000007) % 1000000007;
  }
  printf("%I64d", ans[1]);
  for (int i = 2; i <= n; i++) printf(" %I64d", ans[i]);
  return 0;
}
