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;
template <typename T>
void read(T &x) {
  x = 0;
  char ch = getchar();
  long long fh = 1;
  while (ch < '0' || ch > '9') {
    if (ch == '-') fh = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
  x *= fh;
}
template <typename T>
void write(T x) {
  if (x < 0) x = -x, putchar('-');
  if (x > 9) write(x / 10);
  putchar(x % 10 + '0');
}
template <typename T>
void writeln(T x) {
  write(x);
  puts("");
}
long long qpow(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = res * a % 1000000007;
    a = a * a % 1000000007;
    b >>= 1;
  }
  return res;
}
long long n, k;
long long a[2005];
long long ans[2005];
signed main() {
  read(n);
  read(k);
  for (long long i = 1; i <= n; ++i) read(a[i]);
  if (k == 0) {
    for (long long i = 1; i <= n; ++i) write(a[i]), putchar(' ');
    puts("");
    return 0;
  }
  for (long long i = 1; i <= n; ++i) {
    long long c = 1;
    for (long long j = 1; j <= i; ++j) {
      ans[i] += c * a[i - j + 1] % 1000000007;
      ans[i] %= 1000000007;
      c *= (k + j - 1);
      c %= 1000000007;
      c *= qpow(j, 1000000007 - 2);
      c %= 1000000007;
    }
    write(ans[i]), putchar(' ');
  }
  puts("");
  return 0;
}
