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 maxn = 2000 + 10;
const int mod = 1000000000 + 7;
long long a[maxn], trans[maxn], c[maxn], ret_trans[maxn];
int n;
void mul(long long *A, long long *B) {
  memset(c, 0, sizeof(c));
  for (int i = 0; i < n; i++)
    for (int j = 0; j <= i; j++) c[i] = (c[i] + A[j] * B[i - j]) % mod;
  memcpy(A, c, sizeof(c));
}
int main() {
  int k;
  while (scanf("%d%d", &n, &k) == 2) {
    for (int i = 0; i < n; i++) {
      scanf("%I64d", a + i);
      trans[i] = 1;
    }
    memset(c, 0, sizeof(ret_trans));
    ret_trans[0] = 1;
    while (k) {
      if (k & 1) mul(ret_trans, trans);
      mul(trans, trans);
      k = (k >> 1);
    }
    mul(a, ret_trans);
    for (int i = 0; i < n - 1; i++) printf("%I64d ", a[i]);
    printf("%I64d\n", a[n - 1]);
  }
  return 0;
}
