Digit Division

Digit Division

Time limit: 1 s Memory limit: 512 MiB

  We are given a sequence of n decimal digits. The sequence needs to be partitioned into one or more contiguous subsequences such that each subsequence, when interpreted as a decimal number, is divisible by a given integer m.

  Find the number of different such partitions modulo 109 + 7. When determining if two partitions are different, we only consider the locations of subsequence boundaries rather than the digits themselves, e.g. partitions 2|22 and 22|2 are considered different.

Input

  The first line contains two integers n and m (1 ≤ n ≤ 300 000, 1 ≤ m ≤ 1 000 000) – the length of the sequence and the divisor respectively. The second line contains a string consisting of exactly n digits.

Output

  Output a single integer – the number of different partitions modulo 109 + 7.

Example

input

4 2

1246

output

4

 

 

input

4 7

2015

output

0

 

 

//题意: n 位长的十进制数字，在其中可以任意插入分割线，分割后，要使每一段不为空，并且可以整除 m ，合法分割的方案数

 

//题目是极其简单的，如果前一部分可以整除 m ，那么，这部分乘10的x次方后依然可以整除，然后算出所有可分割的位置后

C(0,all),C(1,all)+...+C(all,all);  这些必然合法

= 2^^all

但是，此题如果没想清楚，写代码会进坑，此题是对方案数取模，all 是%m==0的方案数，进了坑半天想不出来，唉，还是太菜啊，一度wa在第三组，真是日狗了

# include <cstdio>
# include <cstring>
# include <cstdlib>
# include <iostream>
# include <vector>
# include <queue>
# include <stack>
# include <map>
# include <bitset>
# include <sstream>
# include <set>
# include <cmath>
# include <algorithm>
# pragma  comment(linker,"/STACK:102400000,102400000")
using namespace std;
# define LL          long long
# define pr          pair
# define mkp         make_pair
# define lowbit(x)   ((x)&(-x))
# define PI          acos(-1.0)
# define INF         0x3f3f3f3f
# define eps         1e-8
# define MOD         1000000007

inline int scan() {
    int x=0,f=1; char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-') f=-1; ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0'; ch=getchar();}
    return x*f;
}
inline void Out(int a) {
    if(a<0) {putchar('-'); a=-a;}
    if(a>=10) Out(a/10);
    putchar(a%10+'0');
}
# define MX 300050
/**************************/
char num[MX];

LL qk_mi(LL base,LL x)
{
    LL res = 1;
    while (x)
    {
        if (x%2==1) res = (res*base)%MOD;
        base=base*base%MOD;
        x/=2;
    }
    return res;
}

int main()
{
    int n,m;
    while (scanf("%d%d",&n,&m)!=EOF)
    {
        scanf("%s",num);
        LL zuo = 0;
        LL all = 0;
        for (int i=0;i<n;i++)
        {
            zuo=(zuo*10+(num[i]-'0'))%m;
            if (zuo%m==0) all++;
        }
        all--;
        if (zuo%m!=0)
            printf("0\n");
        else
        {
            LL ans = qk_mi(2,all);
            printf("%lld\n",ans);
        }
    }
    return 0;
}