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

6 238 56 497 3 4

output

8

input

8 21 22 3 445 4 3 2

output

32400

Note

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

题意是输入n，k，n表示要得到一串n个数字的号码，k表示要将这段号码分成若干段每一段k个数字，接下来输入n/k个数字，分别表示每一段的数字要是能整出这个数的k位数字，接下来再输入n/k个数字，表示该段数字不能以bi数字开头。

然后，就靠自己的思维去想了，其实是水题。。（求乘方的函数最好自己写，pow如果哪里搞错了我们连查错都无从下手）

#include 
    
     #include 
     
      #include 
      
       #include
       
        #include
        
         #define mod 1000000007using namespace std;typedef long long ll;ll a[100010],b[100010];ll cf(ll x){    ll s=1;    for(int i=0;i
         
          >n>>k; for(ll i=0;i
          
           >a[i]; for(ll i=0;i
           
            >b[i]; ll cc=cf(k)-1,s=1; for(int i=0;i