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 MAX = 2005;
const long long MOD = 1LL * 1000 * 1000 * 1000 + 7;
struct vec {
  int len;
  long long *val;
  vec(int len) : len(len) {
    val = new long long[len];
    for (int i = 0; i < len; i++) val[i] = 0;
  }
  vec() {}
};
vec mult(const vec &a, const vec &b) {
  int len = a.len;
  vec ret(len);
  for (int i = 0; i < len; i++)
    for (int j = 0; j <= i; j++)
      ret.val[i] = (ret.val[i] + a.val[j] * b.val[i - j]) % MOD;
  return ret;
}
vec power(const vec &a, int n) {
  if (n == 1) return a;
  if (n & 1) return mult(a, power(a, n - 1));
  vec ret = power(a, n >> 1);
  return mult(ret, ret);
}
int main() {
  int N, K;
  scanf("%d%d", &N, &K);
  vec a(N);
  for (int i = 0; i < N; i++) scanf("%I64d", a.val + i);
  vec ans;
  if (K) {
    vec one(N);
    for (int i = 0; i < N; i++) one.val[i] = 1;
    ans = power(one, K);
    ans = mult(ans, a);
  } else
    ans = a;
  for (int i = 0; i < N; i++) {
    if (i) printf(" ");
    printf("%I64d", ans.val[i]);
  }
  printf("\n");
  return 0;
}
