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 <iostream>

using namespace std;

int bexp(int a, int x, int p) {
    if (x == 0) {
        return 1;
    }
    if (x % 2 == 1) {
        return a * bexp(a, x - 1, p) % p;
    }
    int t = bexp(a, x / 2, p);
    return t * t % p;
}

int inv(int a, int p) {
    return bexp(a, p - 2, p);
}

int main() {
    int p;
    cin >> p;
    int a[p];
    for (int i = 0; i < p; i++) {
        cin >> a[i];
    }

    int c[p][p];
    for (int i = 0; i < p; i++) {
        c[i][0] = c[i][i] = 1;
        for (int j = 1; j < i; j++) {
            c[i][j] = (c[i-1][j-1] + c[i-1][j]) % p;
        }
    }

    int e[p][p];
    for (int i = 0; i < p; i++) {
        for (int j = 0; j < p; j++) {
            e[i][j] = j == 0 ? 1 : e[i][j-1] * i % p;
        }
    }

    int fac[p];
    for (int i = 0; i < p; i++) {
        fac[i] = i == 0 ? 1 : i * fac[i-1] % p;
    }

    int b[p];
    for (int i = p - 1; i >= 0; i--) {
        int y = 0;
        int neg = 1;
        for (int j = i; j >= 0; j--) {
            y = ((y + a[j] * c[i][j] * neg) % p + p) % p;
            neg = -neg;
        }
        b[i] = y * inv(fac[i], p) % p;

        for (int j = 0; j < p; j++) {
            a[j] = ((a[j] - b[i] * e[j][i]) % p + p) % p;
        }
    }

    for (int i = 0; i < p; i++) {
        cout << b[i] << (i == p - 1 ? '\n' : ' ');
    }

    return 0;
}