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 double PI = acos(-1);
const double eps = 1e-9;
const int inf = 2000000000;
const long long infLL = 9000000000000000000;
inline int in() {
  int x;
  scanf("%d", &x);
  return x;
}
inline long long inl() {
  long long x;
  scanf("%I64d", &x);
  return x;
}
inline double ind() {
  double x;
  scanf("%lf", &x);
  return x;
}
inline bool EQ(double a, double b) { return fabs(a - b) < 1e-9; }
inline bool isLeapYear(long long year) {
  return (year % 400 == 0) || (year % 4 == 0 && year % 100 != 0);
}
inline void normal(long long &a) {
  a %= 1000000007;
  (a < 0) && (a += 1000000007);
}
inline long long modMul(long long a, long long b) {
  a %= 1000000007, b %= 1000000007;
  normal(a), normal(b);
  return (a * b) % 1000000007;
}
inline long long modAdd(long long a, long long b) {
  a %= 1000000007, b %= 1000000007;
  normal(a), normal(b);
  return (a + b) % 1000000007;
}
inline long long modSub(long long a, long long b) {
  a %= 1000000007, b %= 1000000007;
  normal(a), normal(b);
  a -= b;
  normal(a);
  return a;
}
inline long long modPow(long long b, long long p) {
  long long r = 1;
  while (p) {
    if (p & 1) r = modMul(r, b);
    b = modMul(b, b);
    p >>= 1;
  }
  return r;
}
inline long long modInverse(long long a) { return modPow(a, 1000000007 - 2); }
inline long long modDiv(long long a, long long b) {
  return modMul(a, modInverse(b));
}
int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
int power(int x, int n) {
  if (n == 0)
    return 1;
  else if (n % 2 == 0) {
    int y = power(x, n / 2);
    return y * y;
  } else
    return x * power(x, n - 1);
}
long long dp[2005], tmp[2005];
int a[2005];
vector<int> idk;
int main() {
  int n = in(), k = in();
  for (int i = 1; i <= n; i++) {
    a[i] = in();
  }
  if (k == 0) {
    for (int i = 1; i <= n; i++) printf("%d ", a[i]);
    return 0;
  }
  for (int i = 1; i <= n; i++) dp[i] = 1;
  int step = 1;
  int lol = k;
  while (k) {
    if (k & 1) {
      idk.push_back(k);
      k--;
    }
    idk.push_back(k);
    k >>= 1;
  }
  for (int z = idk.size() - 3; z >= 0; z--) {
    if (idk[z] & 1) {
      for (int i = 2; i <= n; i++) dp[i] = modAdd(dp[i], dp[i - 1]);
    } else {
      memset(tmp, 0, sizeof(tmp));
      for (int i = n; i > 1; i--) {
        for (int j = 1; j <= i; j++) {
          tmp[i] = modAdd(tmp[i], modMul(dp[j], dp[i - j + 1]));
        }
      }
      for (int i = 2; i <= n; i++) dp[i] = tmp[i];
      k >>= 1;
    }
  }
  long long tmp = 0;
  for (int i = 1; i <= n; i++) {
    tmp = 0;
    for (int j = 1; j <= i; j++) tmp = modAdd(tmp, modMul(dp[j], a[i - j + 1]));
    printf("%I64d ", tmp);
  }
  printf("\n");
  return 0;
}
