Problem:
Score : 600 points

Problem Statement

Given are a prime number p and a sequence of p integers a_0, \ldots, a_{p-1} consisting of zeros and ones.

Find a polynomial of degree at most p-1, f(x) = b_{p-1} x^{p-1} + b_{p-2} x^{p-2} + \ldots + b_0, satisfying the following conditions:

For each i (0 \leq i \leq p-1), b_i is an integer such that 0 \leq b_i \leq p-1.

For each i (0 \leq i \leq p-1), f(i) \equiv a_i \pmod p.

Constraints

2 \leq p \leq 2999

p is a prime number.

0 \leq a_i \leq 1

Input

Input is given from Standard Input in the following format:

p
a_0 a_1 \ldots a_{p-1}

Output

Print b_0, b_1, \ldots, b_{p-1} of a polynomial f(x) satisfying the conditions, in this order, with spaces in between.

It can be proved that a solution always exists. If multiple solutions exist, any of them will be accepted.

Sample Input 1

2
1 0

Sample Output 1

1 1

f(x) = x + 1 satisfies the conditions, as follows:

f(0) = 0 + 1 = 1 \equiv 1 \pmod 2

f(1) = 1 + 1 = 2 \equiv 0 \pmod 2

Sample Input 2

3
0 0 0

Sample Output 2

0 0 0

f(x) = 0 is also valid.

Sample Input 3

5
0 1 0 1 0

Sample Output 3

0 2 0 1 3

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <iostream>
#include <string>
#include <fstream>
#include <vector>
#include <algorithm>
#include <utility>
#include <set>
#include <queue>
#include <map>
#include <cmath>
#include <numeric>
#include <tuple>

using namespace std;

typedef long long ll;
typedef pair<ll, ll> P;

const ll mod = 1e9 + 7;

long long pow(const long long &v, const long long &n, const long long &mod)
{
    long long ret = 1;
    long long num = n;
    long long now = v;
    
    while (num > 0)
    {
        if (num & 1)
        {
            ret = ret * now % mod;
        }
        now = now * now % mod;
        num = num >> 1;
    }
    
    return ret;
}

ll dp[51][51][2700];

int main() {

    
    ll p;
    cin >> p;
    
    vector<ll> v(p);
    for (int i = 0; i < p; ++i)
    {
        cin >> v[i];
    }
    
    vector<ll> fac = vector<long long>(p + 1, 1);
    vector<ll> inv = vector<long long>(p + 1, 1);
    
    for (int i = 1; i <= p; ++i)
    {
        fac[i] = fac[i - 1] * i % mod;
        inv[i] = pow(fac[i], mod - 2, mod);
    }
    
    vector<ll> vb(p);
    
    for (int i = 0; i < p; ++i)
    {
        if (v[i])
        {
            vb[0] += 1ll;
            vb[0] %= mod;
            for (int j = 0; j < p; ++j)
            {
                if ((!(p % 2) && !(j % 2)) || ((p % 2) && j % 2))
                {
                    vb[j] -= ((fac[p - 1] * inv[j] % mod * inv[p - 1 - j] % mod) * pow(i, p - 1 - j, mod) % mod);
                    vb[j] += mod;
                    vb[j] %= mod;
                }
                else
                {
                    vb[j] += (fac[p - 1] * inv[j] % mod * inv[p - 1 - j] % mod) * pow(i, p - 1 - j, mod) % mod;
                    vb[j] %= mod;
                }
            }
        }
    }
    
    for (int i = 0; i < p; ++i)
    {
        cout << vb[p - 1 - i] << endl;
    }
    
    return 0;
}