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: Time Limit Exceeded)

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

int p;

long long modpow(long long a, long long b){
    if (b == 0) return 1;
    if (b % 2 == 0) return modpow(a, b / 2) * modpow(a, b / 2) % p;
    return a * modpow(a, b - 1) % p;
}

int check(long long x){
    if (x % 2) return p - 1;
    return 1;
}

int main(){
    cin >> p;
    vector<long long> inv(p);
    inv[0] = 0; inv[1] = 1;
    for (int i = 2; i < p; i++) inv[i] = (p - p / i) * inv[p % i] % p;
    vector<long long> C(p);
    C[0] = 1;
    for (int i = 1; i < p; i++) C[i] = (C[i - 1] * (p - i) % p) * inv[i] % p;
    vector<int> a(p);
    for (int i = 0; i < p; i++) cin >> a[i];
    vector<long long> b(p, 0);
    for (int i = 0; i < p; i++){
        if (a[i]){
            b[0] += 1;
            for (int j = 0; j < p; j++) {
                b[j] += (p - (C[j] * modpow(i, p - 1 - j) % p) * check(p - 1 - j) % p) % p;
                b[j] %= p;
            }
        }
    }
    for (int i = 0; i < p; i++){
        cout << b[i];
        if (i == p - 1) cout << endl;
        else cout << " ";
    }
}