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 Kotlin (1.0.0) (judge verdict: Wrong Answer)

import java.util.*

object Mod {
    var num: Long = 1000000007
    val inverse: HashMap<Long, Long> = hashMapOf()
    val factorial: HashMap<Long, Long> = hashMapOf()
}

fun Long.pls(other: Long): Long {
    return (this + other) % Mod.num
}

fun Long.mns(other: Long): Long {
    return (this - other) % Mod.num
}

fun Long.mult(other: Long): Long {
    return ((this % Mod.num) * (other % Mod.num)) % Mod.num
}

fun Long.divi(other: Long): Long {
    return ((this % Mod.num) * other.inverse()) % Mod.num
}

fun Long.power(num: Int): Long {
    var res: Long = 1
    var a = this
    var n = num
    while (n > 0) {
        if (n % 2 == 1) {
            res = res.mult(a)
        }
        a = a.mult(a)
        n = n shr 1
    }
    return res
}

fun Long.inverse(): Long {
    val ans = Mod.inverse[this]
    return if (ans != null) {
        ans
    } else {
        val ans2 = this.power((Mod.num - 2).toInt())
        Mod.inverse[this] = ans2
        ans2
    }
}

fun Long.factorial(): Long {
    if (this == 1.toLong()) return 1
    val ans = Mod.factorial[this]
    return if (ans != null) {
        ans
    } else {
        var ans2: Long = 1
        for (i in 1..this) {
            ans2 = ans2.mult(i)
        }
        Mod.factorial[this] = ans2
        ans2
    }
}

fun Long.combi(other: Long): Long {
    if (this < other) return 0
    if (other == this || other == 0.toLong()) return 1
    return this.factorial()
        .mult(other.factorial().inverse())
        .mult((this - other).factorial().inverse())
}

fun main(args: Array<String>) {
    val p = readLine()!!.toInt()
    Mod.num = p.toLong()
    val array = readLine()!!.split(' ').map(String::toInt)
    val p_1C = hashMapOf<Long, Long>()
    for (k in 0..(p - 1).toLong()) {
        p_1C[k] = (p - 1).toLong().combi(k)
    }
    val ans = Array<Long>(p) { 0 }
    for (i in 0 until p) {
        if (array[i] == 0) continue
        ans[0] = ans[0].pls(1)
        for (k in 0..(p - 1)) {
            ans[k] = ans[k].mns(
                p_1C[k.toLong()]!!.mult((-i.toLong()).power(p - 1 - k))
            )
        }
    }
    println(ans.joinToString(" "))
}