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 = 1e9 + 7;
int add(int x, int y) {
  if ((x += y) >= Mod) x -= Mod;
  return x;
}
int sub(int x, int y) {
  if ((x -= y) < 0) x += Mod;
  return x;
}
int mul(int x, int y) { return (long long)x * y % Mod; }
void add_self(int& x, int y) { x = add(x, y); }
void sub_self(int& x, int y) { x = sub(x, y); }
void mul_self(int& x, int y) { x = mul(x, y); }
int fp(int x, int y) {
  if (!y) return 1;
  if (y & 1) return mul(x, fp(x, y - 1));
  return fp(mul(x, x), y >> 1);
}
int inv(int y) { return fp(y, Mod - 2); }
int C(int n, int r) {
  if (n < 0 || r < 0 || n < r) return 0;
  int Num = 1;
  int Den = 1;
  for (int i = 0; i < r; i++) {
    mul_self(Num, n - i);
    mul_self(Den, i + 1);
  }
  return mul(Num, inv(Den));
}
const int N = 2020;
int a[N];
int b[N];
int main() {
  int n, k;
  cin >> n >> k;
  for (int i = 0; i < n; i++) scanf("%d", a + i);
  b[0] = 1;
  for (int i = 1; i < n; i++) b[i] = C(k + i - 1, i);
  for (int i = 0; i < n; i++) {
    int Ans = 0;
    for (int j = 0; j <= i; j++) add_self(Ans, mul(a[j], b[i - j]));
    printf("%d ", Ans);
  }
}
