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 = int(1e9 + 7);
int n, a[2222], k, c[2222], b[2222];
int modPow(int a, int b) {
  int res = 1;
  while (b > 0) {
    if (b & 1) {
      res = (1LL * a * res) % mod;
    }
    b >>= 1;
    a = (1LL * a * a) % mod;
  }
  return res;
}
void print(int a[]) {
  printf("%d", a[0]);
  for (int i = 1; i < n; ++i) printf(" %d", a[i]);
  puts("");
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 0; i < n; ++i) {
    scanf("%d", &a[i]);
  }
  if (k == 0) {
    print(a);
    return 0;
  }
  --k;
  int cur = 1;
  for (int i = 0; i <= n - 1; ++i) {
    int m = i + k;
    if (i > 0) {
      cur = (1LL * cur * modPow(i, mod - 2)) % mod;
      cur = (1LL * cur * m) % mod;
    }
    c[i] = cur;
  }
  for (int i = 0; i < n; ++i) {
    b[i] = 0;
    for (int j = 0; j <= i; ++j) b[i] = (1LL * c[j] * a[i - j] + b[i]) % mod;
  }
  print(b);
  return 0;
}
