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>
#define rep(i,n) for (int i = 0; i < (int)n; i++)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pi;
typedef pair<pi, pi> pp;
typedef pair<ll, ll> pl;
const double EPS = 1e-9;
const ll MOD = 1000000007;
const int inf = 1 << 30;
const ll linf = 1LL << 60;

ll kai[200001];
ll mokai[200001];
ll p;

ll mod_pow (ll x, ll y) {
    ll ret = 1;
    while (y) {
        if (y&1) ret = ret*x%p;
        x = x*x%p;
        y /= 2;
    }
    return ret;
}

void init (int _n) {
    kai[0] = 1;
    for (int i = 1; i <= _n; i++) kai[i] = kai[i-1]*i%p;
    for (int i = 0; i <= _n; i++) mokai[i] = mod_pow(kai[i],p-2);
}

ll conb (ll x, ll y) {
    ll z = x-y;
    ll ret = kai[x]*mokai[y]%p;
    ret = ret*mokai[z]%p;
    return ret;
}

ll c[3000];
ll prep[3000][3000];

int main() {
    cin >> p;
    init(3000);
    rep(i,p+1) {
        prep[i][0] = 1;
        for (int j = 1; j <= p; j++) prep[i][j] = prep[i][j-1] * i % p;
    }
    rep(i,p) {
        ll a; cin >> a;
        if (a == 0) continue;
        (c[0] += 1) %= p;
        rep(j,p) {
            (c[j] += -1 * conb(p-1, p-1-j) * mod_pow(p-i,p-1-j) + p*p*p) %= p;
        }
    }
    rep(i,p) {
        cout << c[i];
        if (i != p-1) cout << " ";
    }
    cout << endl;
}