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 N;
void isp(const int* T) {
  for (int i = 0; i < N; ++i) printf("%d ", T[i]);
  printf("\n");
}
int tmp[maxn];
int T[maxn];
int sol[maxn];
int X[maxn];
void multy(const int* A, const int* B, int* C) {
  memset(C, 0, maxn * sizeof(int));
  for (int i = 0; i < N; ++i)
    for (int j = 0; j <= i; j++) {
      C[i] = (C[i] + (long long)A[j] * B[i - j]) % mod;
    }
}
void SetSol(int k) {
  for (int i = 0; i < N; i++) T[i] = 1;
  memset(sol, 0, sizeof sol);
  sol[0] = 1;
  for (; k > 0; k >>= 1) {
    if (k & 1) {
      multy(sol, T, tmp);
      for (int i = 0; i < N; i++) sol[i] = tmp[i];
    }
    multy(T, T, tmp);
    for (int i = 0; i < N; ++i) T[i] = tmp[i];
  }
}
int main() {
  int k;
  scanf("%d%d", &N, &k);
  for (int i = 0; i < N; ++i) scanf("%d", X + i);
  SetSol(k);
  multy(X, sol, tmp);
  isp(tmp);
}
