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: Accepted)

#include "bits/stdc++.h"
using namespace std;
typedef long long ll;
const ll MOD = 1e9 + 7;
const ll INF = 1LL << 60;

ll mod_pow(ll a, ll n, ll p) {
	if (n == 0) return 1;
	if (n % 2 == 0) {
		ll tmp = mod_pow(a, n / 2, p);
		return (tmp * tmp) % p;
	}
	return (a * mod_pow(a, n - 1, p)) % p;
}

// 多項式P(x)を受け取って(x-s)P(x)+tという多項式を返す
// 係数はmod p
vector<ll> mul(const vector<ll> &P, int s, int t, ll p) {
	vector<ll> Q(P.size() + 1);
	for (int i = 0; i < P.size(); i++) {
		(Q[i + 1] += P[i]) %= p;
		(Q[i] += P[i] * (p - s) % p) %= p;
	}
	(Q[0] += t) %= p;
	return Q;
}

ll inv[2999];
ll a[2999];
int main() {
	int p;
	cin >> p;
	for (int i = 1; i < p; i++) {
		inv[i] = mod_pow(i, p - 2, p);
	}
	for (int i = 0; i < p; i++) cin >> a[i];
	vector<ll> s, t;
	for (int i = 0; i < p; i++) {
		s.push_back(i);
		t.push_back(a[i]);
		for (int j = i + 1; j < p; j++) {
			(a[j] += p - a[i]) %= p;
			(a[j] *= inv[j - i]) %= p;
		}
	}
	vector<ll> P = { 1 };
	for (int i = p - 1; i >= 0; i--) {
		P = mul(P, s[i], t[i], p);
	}
	(P[1] += P[p]) %= p;
	for (int i = 0; i < p; i++) cout << P[i] << " ";
	cout << endl;
}