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)

#pragma GCC optimize("Ofast")
#include<bits/stdc++.h>
#define int long long
#define gmax(x,y) x=max(x,y)
#define gmin(x,y) x=min(x,y)
#define F first
#define S second
#define P pair
#define FOR(i,a,b) for(int i=a;i<=b;i++)
#define rep(i,a,b) for(int i=a;i<b;i++)
#define V vector
#define RE return
#define ALL(a) a.begin(),a.end()
#define MP make_pair
#define PB emplace_back
#define PF emplace_front
#define FILL(a,b) memset(a,b,sizeof(a))
#define lwb lower_bound
#define upb upper_bound
using namespace std;
int inv[3005],finv[3005],f[3005],mod,n; 
int POW(int x,int y){
	if(!y)RE 1;
	int t=POW(x,y/2);
	t=t*t%mod;
	if(y&1)RE t*x%mod;else RE t;
}
void init(){
	finv[1]=1;f[0]=1;f[1]=1;finv[0]=1;inv[1]=1;
	FOR(i,2,n){
		inv[i]=mod-inv[mod%i]*(mod/i)%mod;
		finv[i]=finv[i-1]*inv[i]%mod;
		f[i]=f[i-1]*i%mod;
	}
}
int C(int x,int y){
	RE f[x]*finv[y]*finv[x-y]; 
}
int ans[3005];
signed main(){
	ios::sync_with_stdio(0);
	cin.tie(0);
	cin>>n;mod=n;
	init();
	bool x;
	int t;
	FOR(i,1,n){
		cin>>x;
		if(!x)continue;
		FOR(j,1,n){
			t=POW(n-i+1,n-j);
			t=t*C(n-1,j-1)%mod;
			ans[j]=(ans[j]-t+mod)%mod;
		}
		ans[1]=(ans[1]+1)%mod;
	}
	FOR(i,1,n)cout<<ans[i]<<' ';
	RE 0;
}