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, ..., a_n / k and b₁, b₂, ..., b_n / k. Let's split the phone number into blocks of length k

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

Pasha asks you to calculate the number of good phone numbers of length n

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, ..., a_n / k (1 ≤ a_i < 10^k).

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

Output

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

Examples

6 2
38 56 49
7 3 4

8

8 2
1 22 3 44
5 4 3 2

32400

自己英語水平捉雞，這道題目看了半天硬是沒看懂

大意就是說有n為一串數字的長度，k為把一串數字分組後每一組中數字的個數
又有a【i】b【i】使得數字串滿足 ，每一組為a【i】的倍數，且不以b【i】開頭；

這是我開始寫的,對於b【i】不能開頭，我是遍歷判斷的.....（不要打我，我只會這個）
結果自然不行

#include<stdio.h>
#include<math.h>
int main(){
    long long undo=1;
    long long a[100000],b[100000],s=0;
    int i,j,n,k,d,mod;
     while(scanf("%d%d",&n,&k)!=EOF){
    d=pow(10,k)-1;mod=pow(10,k-1);
    for(i=0;i<n/k;i++)scanf("%I64d",&a[i]);
    for(i=0;i<n/k;i++)scanf("%I64d",&b[i]);
    for(i=0;i<n/k;i++){
        s+=d/a[i]+1;
        for(j=0;j*a[i]<=d;j++){         //最多9個數量級迴圈判斷，我怕不是個傻子 
            if(j*a[i]/mod==b[i])s--;
             }
             undo*=s;
             undo%=1000000007;
             s=0;
    }
    printf("%I64d\n",undo%1000000007);
    undo=1;
    }
    return 0;
}

　看了大神的程式碼，直接用算式算出不滿足要求2的數字個數，在用總數減去...

又想起新生賽中一道簽到題，等差數列求和，我用了迴圈，答案是直接套公式算，速度肯定快

所以做題時可以直接用算式算出來的，就儘量不要用迴圈獲其他的算，大道至簡

於是改之後...還是wa了 ，而且各種除錯，比較結果都找不出錯

百度之，發現是pow（）的問題，這東西精度不夠，而且裡面都是double型，做題就不要用

函式原型：double pow( double x, double y );

pow

標頭檔案：math.h/cmath(C++中)

功能：計算x的y次方

返回值：x不能為負數且y為小數，或者x為0且y小於等於0，返回冪指數的結果。

ac的程式碼：

 1 #include<stdio.h>
 2 #include<math.h>
 3 long long power(int k){
 4     int s=1;
 5     while(k--)s*=10;
 6     return s;
 7 } 
 8 int main(){
 9     long long undo=1;
10     long long a[100000],b[100000],s=0;
11     int i,j,n,k,d,mod;
12      while(scanf("%d%d",&n,&k)!=EOF){
13          d=power(k); mod=d/10;
14     for(i=0;i<n/k;i++)scanf("%lld",&a[i]);
15     for(i=0;i<n/k;i++)scanf("%lld",&b[i]);
16     for(i=0;i<n/k;i++){
17         s+=(d-1)/a[i]+1;         //帶零，所有a[i]在該位數下的倍數個數 
18         if(b[i]!=0)
19         s-=((b[i]+1)*mod-1)/a[i]-(b[i]*mod-1)/a[i];  //減去10*b[i]~10*（b[i]+1）-1 內ai倍數的個數 eg；30~39                     
20         else if(b[i]==0)
21         s-=(mod-1)/a[i]+1;                   //如果bi是0，減去所有0頭的ai倍數 
22              undo*=s;
23              undo%=1000000007;
24              s=0;
25     }
26     printf("%lld\n",undo);
27     undo=1;
28     }
29     return 0;
30 }

總之：1.能用式子解決的問題，就不要用迴圈獲判斷來解決

           2.能不用pow（），就不用pow（）。