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 Java7 (OpenJDK 1.7.0) (judge verdict: Wrong Answer)

import java.util.*;
import java.io.*;
import java.lang.*;
import java.math.*;
public class Main {
    public static void main(String[] args) throws Exception {
        BufferedReader bf = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
        int n = Integer.parseInt(bf.readLine());
        StringTokenizer st = new StringTokenizer(bf.readLine());
        int[] a = new int[n]; for(int i=0; i<n; i++) a[i] = Integer.parseInt(st.nextToken());
        int p = n;

        long[] ans = new long[n];
        for(int i=1; i<p; i++) {

          for(int j=1; j<p; j++) {
            ans[j] += 1L*(-1*exp(i, p-j, p)+p)*a[i];
            ans[j] += p;
            ans[j] %= p;
          }
        }

          ans[p-1] += 1L*(p-1)*a[0];
          ans[p-1] %= p;
          ans[0] += 1L*a[0];
          ans[0] %= p;
        StringBuilder sb = new StringBuilder();
        for(int i=0; i<n; i++) sb.append(ans[i]+ " ");
        out.println(sb.toString());
        // int n = Integer.parseInt(st.nextToken());

        out.close(); System.exit(0);
    }
    public static int exp(int base, int e, int mod) {
      if(e == 0) return 1;
      if(e == 1) return base;
      int val = exp(base, e/2, mod);
      int ans = (int)(1L*val*val % mod);
      if(e % 2 == 1)
        ans = (int)(1L*ans*base % mod);
      return ans;
    }
    public static int inv(int base, int mod) {
      return exp(base, mod-2, mod);
    }
}