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 mod = 1000000007;
using namespace std;
long long niyuan[2012];
long long exgcd(long long a, long long b, long long& x, long long& y) {
  if (b == 0) {
    x = 1, y = 0;
    return a;
  }
  long long d = exgcd(b, a % b, x, y);
  long long t = x;
  x = y;
  y = t - a / b * y;
  return d;
}
int a[2012];
int main() {
  int i, j, k, t, N, K;
  long long x, y, res, sum;
  for (i = 1; i < 2012; ++i) {
    exgcd(i, mod, x, y);
    if (x < 0) x += mod;
    niyuan[i] = x;
  }
  while (scanf("%d %d", &N, &K) != EOF) {
    for (i = 1; i <= N; ++i) scanf("%d", &a[i]);
    if (K == 0) {
      for (i = 1; i <= N; ++i) {
        printf("%d", a[i]);
        if (i != N) printf(" ");
      }
      printf("\n");
      continue;
    }
    for (i = 1; i <= N; ++i) {
      res = 1;
      sum = 0;
      for (j = 1; j <= i; ++j) {
        sum += res * a[i - j + 1];
        sum %= mod;
        res *= (K + j - 1);
        res %= mod;
        res *= niyuan[j];
        res %= mod;
      }
      printf("%I64d", sum);
      if (i != N) printf(" ");
    }
    printf("\n");
  }
  return 0;
}
