F - Takahashi's Basics in Education and Learning /

### 問題文

この等差数列の初項は A、公差は B です。つまり、s_i = A + B \times i が成り立ちます。

この数列の各項を、先頭に 0 の無い十進法表記に直し、順につなげて読んでできる整数を考えます。たとえば、数列 3, 7, 11, 15, 19 をつなげて読んでできる整数は 37111519 となります。この整数を M で割ったあまりはいくらでしょうか。

### 制約

• 入力はすべて整数である
• 1 \leq L, A, B < 10^{18}
• 2 \leq M \leq 10^9
• 等差数列の要素は全て 10^{18} 未満

### 入力

L A B M


### 入力例 1

5 3 4 10007


### 出力例 1

5563


### 入力例 2

4 8 1 1000000


### 出力例 2

891011


### 入力例 3

107 10000000000007 1000000000000007 998244353


### 出力例 3

39122908


Score : 600 points

### Problem Statement

There is an arithmetic progression with L terms: s_0, s_1, s_2, ... , s_{L-1}.

The initial term is A, and the common difference is B. That is, s_i = A + B \times i holds.

Consider the integer obtained by concatenating the terms written in base ten without leading zeros. For example, the sequence 3, 7, 11, 15, 19 would be concatenated into 37111519. What is the remainder when that integer is divided by M?

### Constraints

• All values in input are integers.
• 1 \leq L, A, B < 10^{18}
• 2 \leq M \leq 10^9
• All terms in the arithmetic progression are less than 10^{18}.

### Input

Input is given from Standard Input in the following format:

L A B M


### Output

Print the remainder when the integer obtained by concatenating the terms is divided by M.

### Sample Input 1

5 3 4 10007


### Sample Output 1

5563


Our arithmetic progression is 3, 7, 11, 15, 19, so the answer is 37111519 mod 10007, that is, 5563.

### Sample Input 2

4 8 1 1000000


### Sample Output 2

891011


### Sample Input 3

107 10000000000007 1000000000000007 998244353


### Sample Output 3

39122908