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>
const long long maxn = 2100;
const long long mo = 1000000007;
long long a[maxn + 1], result[maxn + 1];
long long n, K;
void gcd(long long a, long long b, long long &x, long long &y) {
  long long xpie, ypie;
  if (a % b == 0) {
    x = 0;
    y = 1;
    return;
  }
  gcd(b, a % b, xpie, ypie);
  x = ypie;
  y = xpie - (a / b) * ypie;
}
void init() {
  long long i;
  scanf("%I64d%I64d", &n, &K);
  for (i = 1; i <= n; ++i) {
    scanf("%I64d", &a[i]);
  }
}
long long chu(long long a, long long b) {
  long long x, y;
  gcd(b, mo, x, y);
  if (x < 0) {
    x += (-x) / mo * mo;
    if (x < 0) x += mo;
  } else
    x = x % mo;
  x = (x * a) % mo;
  return x;
}
void work() {
  long long i, j, p, now;
  if (K == 0) {
    for (i = 1; i <= n; ++i) result[i] = a[i];
    return;
  }
  for (i = 1; i <= n; ++i) result[i] = 0;
  now = 1;
  for (i = 1; i <= n; ++i) {
    p = i;
    for (j = 1; j <= n - i + 1; ++j) {
      result[p] += a[j] * now;
      result[p] = result[p] % mo;
      ++p;
    }
    now = now * (K + i - 1) % mo;
    now = chu(now, i);
  }
}
void output() {
  long long i;
  for (i = 1; i <= n; ++i) {
    printf("%I64d", result[i]);
    if (i == n)
      printf("\n");
    else
      printf(" ");
  }
}
int main() {
  init();
  work();
  output();
  return 0;
}
