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>
#define N 3200
using namespace std;
typedef signed long long ll;


ll p;
ll a[N];

ll powmod(ll x) {
    ll y = 1;
    ll n = p - 2;
    while (n) {
        if (n % 2 == 1) {
            y = y * x % p;
        }
        x = x * x % p;
        n >>= 1;
    }
    return y;
}

int main() {
    cin >> p;
    for (auto i = 0; i < p; i++) {
        cin >> a[i];
    }
    
    ll c1[N];
    ll c2[N];
    memset(c1, 0, sizeof(c1));
    memset(c2, 0, sizeof(c2));
    c1[0] = 1;
    for (auto i = 0; i < p; i++) {        
        c2[0] = (-i*c1[0]%p+p)%p;
        for (auto j = 1; j <= i + 1; j++) {
            c2[j] = (c1[j-1]-i*c1[j]%p+p)%p;
        }
                
        swap(c1, c2);
    }
    
    ll c3[N];
    ll ret[N];
    for (auto i = 0; i < p; i++) {
        if (a[i] == 0) {
            continue;
        }
        
        ll q = 1;
        for (auto j = 0; j < p; j++) {
            if (j != i) {
                q = q*(i-j)%p;
            }
        }
        
        q = powmod((q%p+p)%p)%p;
        memset(c3, 0, sizeof(c3));
        for (auto j = p-1; j >= 0; j--) {
            c3[j] = ((c3[j+1]*i+c1[j+1])%p+p)%p;
            ret[j] = (ret[j]+c3[j]*q)%p;
        }
    }
    
    for (auto i = 0; i < p; i++) {
        cout << ret[i] << " ";
    }
    
    return 0;
}