F - Polynomial Construction /

### 問題文

• i (0 \leq i \leq p-1) に対し、b_i0 \leq b_i \leq p-1 なる整数
• i (0 \leq i \leq p-1) に対し、f(i) \equiv a_i \pmod p

### 制約

• 2 \leq p \leq 2999
• p は素数である。
• 0 \leq a_i \leq 1

### 入力

p
a_0 a_1 \ldots a_{p-1}


### 出力

なお、解は必ず存在することが示せる。複数の解が存在する場合は、そのうちのどれを出力してもよい。

### 入力例 1

2
1 0


### 出力例 1

1 1


f(x) = x + 1 は以下のように条件を満たします。

• f(0) = 0 + 1 = 1 \equiv 1 \pmod 2
• f(1) = 1 + 1 = 2 \equiv 0 \pmod 2

### 入力例 2

3
0 0 0


### 出力例 2

0 0 0


f(x) = 0 も有効な出力です。

### 入力例 3

5
0 1 0 1 0


### 出力例 3

0 2 0 1 3


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