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 N = 2009;
const int MOD = 1e9 + 7;
int n;
long long c[N], res[N][N], A[N][N], B[N][N], d[N];
void mul(long long a[N][N], long long b[N][N]) {
  int i, j, k;
  for (i = 0; i < n; ++i) {
    c[i] = 0;
    for (k = 0; k <= i; ++k) {
      c[i] += a[i][k] * b[k][0];
      c[i] %= MOD;
    }
  }
  for (i = 0; i < n; ++i) {
    for (j = 0; j <= i; ++j) {
      a[i][j] = c[abs(i - j)];
    }
  }
}
void modpow(int b) {
  for (int i = 0; i < n; ++i) {
    res[i][i] = 1;
  }
  while (b > 0) {
    if (b & 1) {
      mul(res, A);
    }
    mul(A, A);
    b >>= 1;
  }
}
int main() {
  int k, i, j;
  scanf("%d%d", &n, &k);
  for (i = 0; i < n; ++i) {
    scanf("%d", &B[i][0]);
    B[i][0] %= MOD;
  }
  for (i = 0; i < n; ++i) {
    for (j = 0; j <= i; ++j) {
      A[i][j] = 1;
    }
  }
  if (k == 0) {
    for (i = 0; i < n; ++i) {
      printf("%I64d%c", B[i][0], i == n - 1 ? '\n' : ' ');
      continue;
    }
  } else {
    modpow(k);
    mul(res, B);
    for (i = 0; i < n; ++i) {
      printf("%I64d%c", res[i][0], i == n - 1 ? '\n' : ' ');
    }
  }
  return 0;
}
