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 maxn = 2010;
const int mod = 1e9 + 7;
int a[maxn][maxn];
int ret[maxn][maxn];
int b[maxn];
long long buf[maxn][maxn];
int n, k;
void mul(int a[][maxn], int b[][maxn]) {
  memset(buf, 0, sizeof(buf));
  for (int j = 0; j < n; j++) {
    for (int k = 0; k <= n; k++) {
      buf[0][k] = buf[0][k] + (long long)a[0][j] * b[j][k];
      if (buf[0][k] >= mod) buf[0][k] %= mod;
    }
  }
  for (int i = 1; i < n; i++) {
    for (int j = i; j <= n; j++) {
      buf[i][j] = buf[0][j - i];
    }
  }
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++) {
      a[i][j] = buf[i][j];
    }
}
void get(int k) {
  for (int i = 0; i < n; i++) ret[i][i] = 1;
  while (k) {
    if (k & 1) mul(ret, a);
    mul(a, a);
    k >>= 1;
  }
}
void print(int a[][maxn]) {
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      printf("%d ", a[i][j]);
    }
    puts("");
  }
}
int main() {
  cin >> n >> k;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++) {
      if (i <= j) a[i][j] = 1;
    }
  get(k);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[0][i]);
  }
  mul(a, ret);
  for (int i = 0; i < n; i++) {
    cout << a[0][i] << " ";
  }
  return 0;
}
