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 mod = 1000000000 + 7;
long long fpow(long long x, long long e) {
  long long res = 1;
  while (e) {
    if (e & 1) {
      res *= x;
      res %= mod;
    }
    x *= x;
    x %= mod;
    e >>= 1;
  }
  return res;
}
long long mydiv(long long x, long long y) {
  long long res = x * fpow(y, mod - 2);
  return res % mod;
}
long long getC(long long a, long long b) {
  long long res = 1;
  for (long long i = 1; i <= b; ++i) {
    res *= (a - b + i);
    res %= mod;
    res = mydiv(res, i);
  }
  return res;
}
const int N = 2048;
int ar[N];
long long res[N];
int n, k;
long long C[N];
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 0; i < n; ++i) scanf("%d", ar + i);
  if (k == 0) {
    for (int i = 0; i < n; ++i) printf("%d ", ar[i]);
    printf("\n");
    return 0;
  }
  for (int i = 0; i < n; ++i) {
    C[i] = getC(k - 1 + i, i);
  }
  for (int i = 0; i < n; ++i) {
    res[i] = 0;
    for (int j = 0; j <= i; ++j) {
      res[i] += ar[j] * C[i - j];
      res[i] %= mod;
    }
  }
  for (int i = 0; i < n; ++i) printf("%d ", int(res[i]));
  printf("\n");
  return 0;
}
