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 N = 2005;
int n, k, a[N], l;
long long m[N];
int calc[100];
long long t[N];
void multi1() {
  int i, j;
  for (i = 0; i < n; i++) {
    long long tmp = 0;
    for (j = 0; j <= i; j++) {
      tmp += m[i - j] * m[j];
      tmp %= 1000000007;
    }
    t[i] = tmp;
  }
  for (i = 0; i < n; i++) m[i] = t[i];
}
void multi2() {
  int i, j;
  for (i = 0; i < n; i++) {
    long long tmp = 0;
    for (j = 0; j <= i; j++) {
      tmp += m[i - j];
      tmp %= 1000000007;
    }
    t[i] = tmp;
  }
  for (i = 0; i < n; i++) m[i] = t[i];
}
void quickpow() {
  int i, j, t = k;
  l = 0;
  while (t) {
    calc[l++] = t % 2;
    t /= 2;
  }
  for (i = l - 1; i >= 0; i--) {
    multi1();
    if (calc[i]) {
      multi2();
    }
  }
}
int main() {
  scanf("%d%d", &n, &k);
  int i, j;
  for (i = 0; i < n; i++) scanf("%d", &a[i]);
  memset(m, 0, sizeof(m));
  m[0] = 1;
  quickpow();
  for (i = 0; i < n; i++) {
    long long ans = 0;
    for (j = 0; j <= i; j++) {
      ans += (long long)a[j] * (long long)m[i - j];
      ans %= 1000000007;
    }
    printf("%d", (int)ans);
    if (i == n - 1)
      printf("\n");
    else
      printf(" ");
  }
  return 0;
}
