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;
bool workingLocal = false;
const int INF = 1000 * 1000 * 1000 + 7;
const double EPS = 1e-9;
template <class T>
void chmin(T &t, T f) {
  if (t > f) t = f;
}
template <class T>
void chmax(T &t, T f) {
  if (t < f) t = f;
}
inline int getint() {
  int a;
  return scanf("%d", &a) ? a : (fprintf(stderr, "trying to read\n"), -1);
}
inline double getdouble() {
  double a;
  return scanf("%lf", &a) ? a : (fprintf(stderr, "trying to read\n"), -1);
}
long long ApowBmodC(long long a, int b, int c) {
  if (b == 0) return 1 - (c == 0);
  long long ret = ApowBmodC(a * a % c, b / 2, c);
  if (b & 1) ret = ret * a % c;
  return ret;
}
int inv(int first) { return ApowBmodC(first, INF - 2, INF); }
int nCk(int n, int k) {
  long long ret = 1;
  if (k == 0) return 1;
  chmin(k, n - k);
  if (k < 0 || k > n) return 0;
  for (int i = (int)0; i < (int)k; i++) {
    ret *= n - i;
    ret %= INF;
    ret *= inv(i + 1);
    ret %= INF;
  }
  return ret;
}
const int N = 5e3;
long long a[N], nck[N], sum[N];
void myCode() {
  int n = getint(), k = getint();
  for (int i = (int)0; i < (int)n; i++) a[i] = getint();
  for (int i = (int)0; i < (int)n; i++) nck[i] = nCk(k - 1 + i, i);
  for (int i = (int)0; i < (int)n; i++) {
    int ret = 0;
    for (int j = (int)0; j < (int)i + 1; j++)
      ret = (ret + a[i - j] * nck[j]) % INF;
    printf("%d%c", ret, (i == n - 1) ? '\n' : ' ');
  }
}
int main() {
  myCode();
  return 0;
}
