Codeforces Round #326 (Div. 2) B. Pasha and Phone C. Duff and Weight Lifting

　　　　　　　　　　　　　　　　　　　　　　　　B. Pasha and Phone

Pasha has recently bought a new phone jPager and started adding his friends' phone numbers there. Each phone number consists of exactly n digits.

Also Pasha has a number k and two sequences of length n / k (n is divisible by k) a1, a2, ..., an / k and b1, b2, ..., bn / k. Let's split the phone number into blocks of length k. The first block will be formed by digits from the phone number that are on positions 1, 2,..., k, the second block will be formed by digits from the phone number that are on positions k + 1, k + 2, ..., 2·k and so on. Pasha considers a phone number good, if the i-th block doesn't start from the digit bi and is divisible by ai if represented as an integer.

To represent the block of length k as an integer, let's write it out as a sequence c1, c2,...,ck. Then the integer is calculated as the result of the expression c1·10k - 1 + c2·10k - 2 + ... + ck.

Pasha asks you to calculate the number of good phone numbers of length n, for the given k, ai and bi. As this number can be too big, print it modulo 109 + 7.

Input

The first line of the input contains two integers n and k (1 ≤ n ≤ 100 000, 1 ≤ k ≤ min(n, 9)) — the length of all phone numbers and the length of each block, respectively. It is guaranteed that n is divisible by k.

The second line of the input contains n / k space-separated positive integers — sequence a1, a2, ..., an / k (1 ≤ ai < 10k).

The third line of the input contains n / k space-separated positive integers — sequence b1, b2, ..., bn / k (0 ≤ bi ≤ 9).

Output

Print a single integer — the number of good phone numbers of length n modulo 109 + 7.

Sample test(s)

input

6 2

38 56 49

7 3 4

output

8

8 2

1 22 3 44

5 4 3 2

32400

Note

In the first test sample good phone numbers are: 000000, 000098, 005600, 005698, 380000, 380098, 385600, 385698.

題意：将一個長度為n的phone number，分成長度為k的n/k給塊，如果任意一個塊(塊i)中的數字x的開頭數字不是b[i], 且x可以整除a[i]，那麼就稱這個phone number是 good number。問這樣的good number 有多少個！

思路：容斥原理。 塊 i : tot = 數字長度為k且可以整除a[i]的個數；

 bi_tot = 數字長度為k且開頭數字為b[i]且可以整除a[i]的個數 - 數字長度為k且開頭數字為(b[i]-1)且可以整除a[i]的個數

那麼符合要求的數字個數 = tot - bi_tot;(b[i]>0)

　　　　　　　　　　　　　　　　　　　　　　　　C. Duff and Weight Lifting

Recently, Duff has been practicing weight lifting. As a hard practice, Malek gave her a task. He gave her a sequence of weights. Weight of i-th of them is 2^wi pounds. In each step, Duff can lift some of the remaining weights and throw them away. She does this until there's no more weight left. Malek asked her to minimize the number of steps.

Duff is a competitive programming fan. That's why in each step, she can only lift and throw away a sequence of weights 2^a1, ..., 2^ak if and only if there exists a non-negative integer x such that 2^a1 + 2^a2 + ... + 2^ak = 2^x, i. e. the sum of those numbers is a power of two.

Duff is a competitive programming fan, but not a programmer. That's why she asked for your help. Help her minimize the number of steps.

The first line of input contains integer n (1 ≤ n ≤ 106), the number of weights.

The second line contains n integers w1, ..., wn separated by spaces (0 ≤ wi ≤ 106 for each 1 ≤ i ≤ n), the powers of two forming the weights values.

Print the minimum number of steps in a single line.

In the first sample case: One optimal way would be to throw away the first three in the first step and the rest in the second step. Also, it's not possible to do it in one step because their sum is not a power of two.

In the second sample case: The only optimal way is to throw away one weight in each step. It's not possible to do it in less than 4 steps because there's no subset of weights with more than one weight and sum equal to a power of two.

題意：給定n個數w1, w2, w3,.......wn, 然後從這個n數中找出這樣的一個子序列a1,a2,a3....ak 使得 2^a1+2^a2.....+2^ak = 2^x, 然後可以删除這個序列，

那麼最少經過幾步可以全部将這n個數删除！

思路：其實就是 二進制數 相加的過程，n個數對應n個二進制數，從最低位到最高位相加，得到最後的二進制數中 1 的個數就是答案！