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 long long mod = 1000000007;
long long a[2010];
long long POW(long long a, long long b) {
  long long ret = 1;
  while (b) {
    if (b & 1) ret = ret * a % mod;
    a = a * a % mod;
    b >>= 1;
  }
  return ret;
}
long long inv[2010];
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%I64d", &a[i]);
  if (k == 0) {
    for (int i = 1; i <= n; i++) printf("%d ", a[i]);
    return 0;
  }
  long long fac = 1;
  inv[0] = 1;
  for (int i = 1; i <= n; i++) {
    fac *= i;
    fac %= mod;
    inv[i] = POW(fac, mod - 2);
  }
  for (int i = 1; i <= n; i++) {
    long long ans = 0;
    int x = i;
    long long up = 1, down = 1;
    int ff = k - 1;
    for (int j = 0; j < i; j++) {
      if (ff && j) up = up * ff % mod;
      if (j) down *= (j);
      down %= mod;
      long long tmp = 1;
      if (ff > 0) tmp = up * inv[j] % mod;
      ans += tmp * a[x] % mod;
      x--;
      ff++;
      if (ans > mod) ans -= mod;
    }
    printf("%I64d ", ans);
  }
  return 0;
}
