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 + 5, P = 1000000007;
int N, K, a[MAXN];
long long b[MAXN], cn[MAXN];
long long FastPow(long long a, int p, int m) {
  long long ret = 1;
  while (p) {
    if (p & 1) ret = ret * a % m;
    if (p >>= 1) a = a * a % m;
  }
  return ret;
}
int main() {
  while (~scanf("%d %d", &N, &K)) {
    for (int i = 0; i < N; i++) scanf("%d", a + i);
    for (int j = 1; j < N; j++) {
      for (int i = N - 1; i >= j; i--) {
        a[i] = ((a[i] - a[i - 1]) % P + P) % P;
      }
    }
    cn[0] = 1;
    for (int i = 1; i < N; i++)
      if (i > K)
        cn[i] = 0;
      else
        cn[i] = cn[i - 1] * (K - i + 1) % P * FastPow(i, P - 2, P) % P;
    memset(b, 0, sizeof(b));
    for (int i = 0; i < N; i++) {
      for (int j = 0; j <= i; j++) b[i] = (b[i] + cn[j] * a[i - j]) % P;
      for (int i = N - 1; i > 0; i--) cn[i] = (cn[i] + cn[i - 1]) % P;
    }
    for (int i = 0; i < N; i++) {
      printf("%d%c", (int)b[i], i < N - 1 ? ' ' : '\n');
    }
  }
  return 0;
}
