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>
 
using namespace std;
 
#define DEBUG(x) cerr<<#x<<": "<<x<<endl;
#define DEBUG_VEC(v) cerr<<#v<<":";for(int i=0;i<v.size();i++) cerr<<" "<<v[i]; cerr<<endl;
#define DEBUG_MAT(v) cerr<<#v<<endl;for(int i=0;i<v.size();i++){for(int j=0;j<v[i].size();j++) {cerr<<v[i][j]<<" ";}cerr<<endl;}
typedef long long ll;
#define int ll
 
#define vi vector<int>
#define vl vector<ll>
#define vii vector< vector<int> >
#define vll vector< vector<ll> >
#define vs vector<string>
#define pii pair<int,int>
#define pis pair<int,string>
#define psi pair<string,int>
#define pll pair<ll,ll>
template<class S, class T> pair<S, T> operator+(const pair<S, T> &s, const pair<S, T> &t) { return pair<S, T>(s.first + t.first, s.second + t.second); }
template<class S, class T> pair<S, T> operator-(const pair<S, T> &s, const pair<S, T> &t) { return pair<S, T>(s.first - t.first, s.second - t.second); }
template<class S, class T> ostream& operator<<(ostream& os, pair<S, T> p) { os << "(" << p.first << ", " << p.second << ")"; return os; }
#define X first
#define Y second
#define rep(i,n) for(int i=0;i<(int)(n);i++)
#define rep1(i,n) for(int i=1;i<=(int)(n);i++)
#define rrep(i,n) for(int i=(int)(n)-1;i>=0;i--)
#define rrep1(i,n) for(int i=(int)(n);i>0;i--)
#define REP(i,a,b) for(int i=a;i<b;i++)
#define in(x, a, b) (a <= x && x < b)
#define all(c) c.begin(),c.end()
template<class T> bool chmax(T &a, const T &b) { if (a<b) { a = b; return 1; } return 0; }
template<class T> bool chmin(T &a, const T &b) { if (a>b) { a = b; return 1; } return 0; }
#define UNIQUE(v) v.erase(std::unique(v.begin(), v.end()), v.end());
const ll inf = 1000000001;
const ll INF = (ll)1e18 + 1;
const long double pi = 3.1415926535897932384626433832795028841971L;
#define Sp(p) cout<<setprecision(25)<< fixed<<p<<endl;
//int dx[4] = {1, 0, -1, 0}, dy[4] = {0, 1, 0, -1};
//int dx2[8] = { 1,1,0,-1,-1,-1,0,1 }, dy2[8] = { 0,1,1,1,0,-1,-1,-1 };
vi dx = {1, 0, -1, 0}, dy = {0, 1, 0, -1};
vi dx2 = { 1,1,0,-1,-1,-1,0,1 }, dy2 = { 0,1,1,1,0,-1,-1,-1 };
#define fio() cin.tie(0); ios::sync_with_stdio(false);
const ll MOD = 1000000007;
//const ll MOD = 998244353;
// #define mp make_pair
//#define endl '\n'

const int MAXN = 555555;

vl fact(MAXN);
vl rfact(MAXN);

ll mod_pow(ll x, ll p, ll M = MOD) {
  if (p < 0) {
    x = mod_pow(x, M - 2, M);
    p = -p;
  }
  ll a = 1;
  while (p) {
    if (p % 2)
      a = a*x%M;
    x = x*x%M;
    p /= 2;
  }
  return a;
}

ll mod_inverse(ll a, ll M = MOD) {
  return mod_pow(a, M - 2, M);
}

void set_fact(ll n, ll M = MOD) {
  fact[0] = fact[1] = rfact[0] = rfact[1] = 1;
  for (ll i = 2; i <= n; i++) {
    fact[i] = i * fact[i - 1] % M;
    // rfact[i] = mod_inverse(fact[i], M);
  }
}

//http://drken1215.hatenablog.com/entry/2018/06/08/210000
//n���傫��fact���v�Z�ł��Ȃ��Ƃ��̂ق��̌v�Z���@�ɂ��ď����Ă���
ll nCr(ll n, ll r, ll M = MOD) {
  if (r > n) return 0;
  assert(fact[2] == 2);
  ll ret = fact[n];
  if (rfact[r] == 0) {
    rfact[r] = mod_inverse(fact[r], M);
  }
  ret = (ret*rfact[r]) % M;
  if (rfact[n - r] == 0) {
    rfact[n - r] = mod_inverse(fact[n - r], M);
  }
  ret = (ret*rfact[n - r]) % M;
  return ret;
}

ll nHr(ll n, ll r) {
  return nCr(n+r-1, r);
}

vl kei;
int p;

vl calc(int k) {
    vl res = kei;
    for (int i = 1; i < p; i++) {
        (res[i] -= res[i - 1] * k % p) %= p;
        //DEBUG_VEC(res);
    }
    return res;
}

signed main() {
    fio();
    cin >> p;
    vi a(p);
    rep (i, p) cin >> a[i];

    kei.resize(p + 1);
    kei[0] = 1;
    for (int i = 0; i < p; i++) {
        for (int j = i + 1; j > 0; j--) {
            (kei[j] += kei[j - 1] * i % p) %= p;
        }
    }

    vl inv_memo(p);
    for (int i = 1; i < p; i++) inv_memo[i] = mod_inverse(i, p);

    // DEBUG_VEC(kei);
    // rep (i, p) {
    //     vl aaa = calc(i);
    //     DEBUG_VEC(aaa);
    // }
    
    vl ans(p);
    rep (i, p) {
        if (a[i] == 0) continue;
        int bunbo = 1;
        rep (j, p) {
            if (j == i) continue;
            bunbo = bunbo * (i - j) % p;
        }
        vl bunsi = calc(i);
        for (int i = 0; i < p; i++) {
            if (i % 2 == 0) ans[i] += bunsi[i] * inv_memo[bunbo] % p;
            else ans[i] -= bunsi[i] * inv_memo[bunbo] % p;
        }
    }

    rrep (i, p) {
        ans[i] = (ans[i] % p + p) % p;
        cout << ans[i] << " ";
    }
    cout << endl;
}