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 long long MOD = (int)1e9 + 7;
struct vct {
  long long v[2010];
} b, a;
int c[2010];
int n, k;
vct mul(vct x, vct y) {
  vct res;
  memset(res.v, 0LL, sizeof(res.v));
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= i; j++) {
      res.v[i] = (res.v[i] + x.v[j] * y.v[i - j + 1]) % MOD;
    }
  }
  return res;
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%d", &c[i]);
  for (int i = 1; i <= n; i++) b.v[i] = 1LL;
  memset(a.v, 0LL, sizeof(a.v));
  a.v[1] = 1LL;
  while (k > 0) {
    if (k & 1) a = mul(a, b);
    vct temp = mul(b, b);
    b = temp;
    k >>= 1;
  }
  long long ans;
  for (int i = 1; i <= n; i++) {
    ans = 0LL;
    for (int j = i; j >= 1; j--) {
      ans = (ans + a.v[j] * (long long)c[i - j + 1]) % MOD;
    }
    printf("%d ", ans);
  }
  printf("\n");
  return 0;
}
