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 <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <set>
#include <map>

#define inf 0x3f3f3f3f
#define Inf 0x3FFFFFFFFFFFFFFFLL
#define maxn 2020
using namespace std;
typedef __int64 ll;
ll const mod = 1000000007LL;
ll arr[maxn];

ll fac[maxn], rev[maxn];
ll powMod(ll a, ll b) {
    ll ret = 1LL;
    while (b) {
        if (b & 1) ret = ret * a % mod;
        a = a * a % mod;
        b >>= 1;
    }
    return ret;
}
void Cinit() {
    fac[0] = rev[0] = 1LL;
    for (int i = 1; i <= 2000; ++i) {
        fac[i] = fac[i - 1] * i % mod;
        rev[i] = powMod(fac[i], mod - 2);
    }
}
ll C(ll n, ll m) {
    ll ret = rev[m];
    for (ll i = 0; i < m; ++i) {
        ret = ret * (n - i) % mod;
    }
    return ret;
}
ll mul[maxn];

int main() {
    Cinit();
    int n, k;
//    while(cin>>n>>k)cout<<C(n,k)<<endl;
    while (~scanf("%d%d", &n, &k)) {
        for (int i = 0; i < n; ++i) {
            scanf("%I64d", &arr[i]);
        }
        if (k == 0) {
            for (int i = 0; i < n; ++i) {
                if (i == n - 1) printf("%I64d\n", arr[i]);
                else printf("%I64d ", arr[i]);
            }
            continue;
        }
        ll ret = 0LL;
        for (int i = 0; i < n; ++i) {
            mul[i] = C(k + i - 1, i) % mod;
//            printf("%I64d ", mul[i]);
//            if(i==n-1)puts("");
        }
        for (int i = 0; i < n; ++i) {
            ll t = 0;
            for (int j = 0; j <= i; ++j) {
                t += arr[j] * mul[i - j];
            }
            if (i == n - 1) printf("%I64d\n", t);
            else printf("%I64d ", t);
        }
    }
    return 0;
}
