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 (Clang 3.8.0) (judge verdict: Time Limit Exceeded)

#include <algorithm>
#include <bitset>
#include <complex>
#include <deque>
#include <exception>
#include <fstream>
#include <functional>
#include <iomanip>
#include <ios>
#include <iosfwd>
#include <iostream>
#include <istream>
#include <iterator>
#include <limits>
#include <list>
#include <locale>
#include <map>
#include <memory>
#include <new>
#include <numeric>
#include <ostream>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <stdexcept>
#include <streambuf>
#include <string>
#include <typeinfo>
#include <utility>
#include <valarray>
#include <vector>
#include <climits>
#include <cstring>

#define rep(i, m, n) for(int i=int(m);i<int(n);i++)
#define all(c) begin(c),end(c)

template<typename T1, typename T2>
inline void chmin(T1 &a, T2 b) { if (a > b) a = b; }

template<typename T1, typename T2>
inline void chmax(T1 &a, T2 b) { if (a < b) a = b; }

//改造
typedef long long int ll;
using ll = long long int;
using ull = long long unsigned int;
using namespace std;
#define INF (1 << 30) - 1
#define INFl (ll)5e15
#define DEBUG 0 //デバッグする時1にしてね
#define dump(x)  cerr << #x << " = " << (x) << endl
ll MOD = 0ll;


//ここから編集する
class Combination {

public:
    vector<ll> fact;

    Combination(int max_n) {
        //1 ~ n!までの階乗modを求める
        int n = max_n;
        vector<ll> fact(n + 1);
        fact[0] = 1;
        for (int i = 1; i <= n; i++) {
            fact[i] = (fact[i - 1] * i) % MOD;
        }
        this->fact = fact;
    }

    template<typename T>
    //templateをつかってみｔか
    ll pow(T x, int y) {
        if (y == 0) return 1;
        if (y == 1) return x;
        if (y == 2) return (x * x) % MOD;
//        if (y % 2 == 1) return pow(x, y / 2) * pow(x, y / 2) % MOD;
        if (y % 2 == 1) return pow(pow(x, y / 2), 2) * x % MOD;
//        return pow(x, y / 2) * pow(x, y / 2) % MOD;
        return pow(pow(x, y / 2), 2) % MOD;
    }

    ll nCk(int n, int k) {
//        return (fact[n] * pow(fact[n-k],MOD-2) * pow(fact[k],MOD-2)) % MOD;
        return (((fact[n] * pow(fact[n - k], MOD - 2)) % MOD) * pow(fact[k], MOD - 2)) % MOD;
    }

    /**
     * x ^ -1を求める
     */
    template<typename T>
    ll inv(T x) {
        return pow(x, MOD - 2);
    }
};

class Solve {
public:
    void solve() {
        int p;
        cin >> p;
        vector<int> b(p);
        MOD = p;
        Combination cmb(p);

        vector<vector<ll>> nck(p, vector<ll>(p));
        nck[1][0] = nck[1][1] = 1;
        for (int i = 2; i < p; ++i) {
            for (int j = 0; j < p; ++j) {
                if (j == 0 || j == i) {
                    nck[i][j] = 1;
                } else {
                    nck[i][j] = (nck[i - 1][j - 1] + nck[i - 1][j]) % MOD;
                }
            }
        }

        for (int i = 0; i < p; ++i) {
            int a;
            cin >> a;
            if (a == 1) {
                b[0]++;
                for (int k = 0; k < p; ++k) {
//                    ll tmp = cmb.nCk(p - 1, k) * cmb.pow(-i, p - 1 - k) % MOD;
                    ll tmp = nck[p - 1][k] * cmb.pow(-i, p - 1 - k) % MOD;
                    b[k] -= tmp;
                    b[k] %= MOD;
                    if (b[k] < 0) b[k] += MOD;
                }
            }
        }

        for (int i = 0; i < p; ++i) {
            cout << b[i] << " ";
        }
        cout << endl;
    }
};


int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    cout << fixed << setprecision(10);

    Solve().solve();


    return 0;
}