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 <bits/stdc++.h>
using namespace std;

const int P = 3005;

int p, u;
bool neg;
vector<int> fi = {1}, se = {1};

vector<int> operator*(vector<int> &a, vector<int> &b)
{
    int sz = min((int)a.size() + (int)b.size() - 1, p - 1);
    cout << sz << '\n';
    vector<int> ans(sz, 0);
    for (int i = 0; i < (int)a.size(); i++)
        for (int j = 0; j < (int)b.size(); j++)
            (ans[(i + j) % (p - 1)] += a[i] * b[j]) %= p;
    return ans;
}

vector<int> operator^(vector<int> cur, int p)
{
    if (p == 1)
        return cur;
    vector<int> tmp = cur ^ (p / 2);
    tmp = tmp * tmp;
    if (p & 1)
        tmp = tmp * cur;
    return tmp;
}

int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    cin >> p >> u;
    neg = (u == 1);
    for (int i = 1; i < p; i++)
    {
        cin >> u;
        vector<int> cur = {p - i, 1};
        if (u == neg)
            fi = fi * cur;
        else
            se = se * cur;
    }
    (++se[0]) %= p;
    fi = fi ^ (p - 1);
    fi = fi * se;
    if (neg)
        for (int &v : fi)
            v = (p - v) % p;
    fi.push_back(fi[0]);
    fi[0] = neg;
    for (int &v : fi)
        cout << v << " ";
}