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 int p = 1000000007;
using namespace std;
long long m[2001], ans[2001], tmp[2001];
int n;
bool f;
void m1() {
  if (f) {
    for (int i = 1; i <= n; i++) ans[i] = m[i];
    f = false;
    return;
  }
  int i, j;
  for (i = 1; i <= n; i++) {
    int s = 0;
    for (j = 1; j <= i; j++) s = (s + 1LL * m[j] * ans[i + 1 - j] % p) % p;
    tmp[i] = s;
  }
  for (i = 1; i <= n; i++) ans[i] = tmp[i];
}
void m2() {
  int i, j;
  for (i = 1; i <= n; i++) {
    int s = 0;
    for (j = 1; j <= i; j++) s = (s + 1LL * m[j] * m[i + 1 - j] % p) % p;
    tmp[i] = s;
  }
  for (i = 1; i <= n; i++) m[i] = tmp[i];
}
int main() {
  int i, j, k;
  long long a[2001];
  scanf("%d%d", &n, &k);
  for (i = 1; i <= n; i++) {
    scanf("%I64d", &a[i]);
    if (k == 0) printf("%I64d ", a[i]);
    m[i] = 1;
  }
  if (k == 0) return 0;
  f = true;
  while (k) {
    if (k & 1) m1();
    k = k >> 1;
    m2();
  }
  long long s = 0;
  for (i = 1; i <= n; i++) {
    s = 0;
    for (j = 1; j <= i; j++) s = (s + 1LL * a[j] * ans[i + 1 - j] % p) % p;
    printf("%I64d ", s);
  }
  return 0;
}
