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 = 1000000007;
const int N = 2000;
int n;
inline int times(int a, int b) { return (int)(((long long int)a * b) % mod); }
void conv(int a[N], int b[N], int c[N]) {
  for (int k = 0; k < n; k++) {
    c[k] = 0;
    for (int i = 0; i <= k; i++) {
      c[k] += times(a[i], b[k - i]);
      c[k] %= mod;
    }
  }
}
void mult(int a[N], int x[N], int res[N]) {
  for (int j = 0; j < n; j++) {
    res[j] = 0;
    for (int i = 0; j + i < n; i++) {
      res[j] += times(a[i], x[j + i]);
      res[j] %= mod;
    }
  }
}
void exp_mat(int a[N], int exp, int res[N]) {
  if (exp == 0) {
    for (int i = 0; i < n; i++) res[i] = 0;
    res[0] = 1 % mod;
  } else {
    int tmp[N];
    int halfsq[N];
    exp_mat(a, exp / 2, tmp);
    conv(tmp, tmp, halfsq);
    if (exp % 2 == 0) {
      for (int i = 0; i < n; i++) tmp[i] = 0;
      tmp[0] = 1 % mod;
    } else {
      for (int i = 0; i < n; i++) tmp[i] = a[i];
    }
    conv(tmp, halfsq, res);
  }
}
int main() {
  int k;
  int x[N];
  scanf("%d", &n);
  scanf("%d", &k);
  for (int i = 0; i < n; i++) scanf("%d", &x[i]);
  reverse(x, x + n);
  int a[N];
  int exxp[N];
  int res[N];
  for (int i = 0; i < n; i++) a[i] = 1;
  exp_mat(a, k, exxp);
  mult(exxp, x, res);
  reverse(res, res + n);
  for (int i = 0; i < n; i++) printf("%d ", res[i]);
  printf("\n");
  return 0;
}
