CodeForces - 628D （數位dp）

Consider the decimal presentation of an integer. Let's call a number d-magic if digit d appears in decimal presentation of the number on even positions and nowhere else.

For example, the numbers 1727374, 17, 1 are 7-magic but 77, 7, 123, 34, 71 are not 7-magic. On the other hand the number 7 is 0-magic, 123 is 2-magic, 34 is 4-magicand 71 is 1-magic.

Find the number of d-magic numbers in the segment [a, b] that are multiple of m. Because the answer can be very huge you should only find its value modulo 109 + 7 (so you should find the remainder after dividing by 109 + 7).

Input

The first line contains two integers m, d (1 ≤ m ≤ 2000, 0 ≤ d ≤ 9) — the parameters from the problem statement.

The second line contains positive integer a in decimal presentation (without leading zeroes).

The third line contains positive integer b in decimal presentation (without leading zeroes).

It is guaranteed that a ≤ b, the number of digits in a and b are the same and don't exceed 2000.

Output

Print the only integer a — the remainder after dividing by 109 + 7 of the number of d-magic numbers in segment [a, b] that are multiple of m.

Examples

Input

2 6
10
99

Output

8

Input

2 0
1
9

Output

4

Input

19 7
1000
9999

Output

6

 

題目大意：

求一個區間內某種數的個數，需要是m的倍數，並且奇數位不能包含d，偶數為必須包含d。數的位數是2000位。

思路：

首先是數位dp。

關於m的限制，直接便計算邊取模就行。d限制，需要分奇偶討論一下。

數的位數很大，需要用字串，這樣的話，閉區間，左邊的數減一不好計算，需要特判一下。

程式碼：

 

#include <cstdio>
#include <iostream>
#include <cstring>
#include <vector>
#include <cmath>
#include <algorithm>
#include <map>
#include <queue>
#include <cmath>
#include <ctime>

using namespace std;
#define ll long long
const int mod=1e9+7;
long long a[2200];
long long dp[2200][2100][2];
int m, id;
int l;
long long dfs(int pos, int sum, int jo,int limit)
{
	if (pos == -1)
	{
	    return !sum&&jo;
	}
	if (!limit&&dp[pos][sum][jo] != -1)return dp[pos][sum][jo];
	long long sun = 0;
	int end = limit ? a[pos] : 9;
	for (int i = 0;i <= end;i++)
	{
		if ((l-pos)%2==0)
		{

            sun = (sun+dfs(pos - 1, (sum*10 + i) % m,jo&&(i==id), limit&&i == a[pos]))%mod;

		}
		else
		{
			sun = (sun+dfs(pos - 1, (sum*10 + i) % m, (jo&&!(i==id)),limit&&i == a[pos]))%mod;
		}

	}
	if (!limit)dp[pos][sum][jo] = sun;
	return sun;
}

long  long go(char x[])
{

	int pos = strlen(x);
	l=pos;
	memset(a,0,sizeof(a));
	for(int i=0;i<pos;i++)
	{
	    a[pos-i-1]=x[i]-'0';
	}
	return dfs(pos - 1, 0, 1,1);
}

char b[2200],c[2200];
int pan(char x[])
{
    int pos = strlen(x);
    int sum=0;
    int flag=0;
    int flag1=0;
    for(int i=0;i<pos;i++)
    {
        int v=x[i]-'0';
        sum=(sum*10+v)%m;
        if(i%2==1)
        {
            if(v!=id)
            {
                flag=1;
                break;
            }
        }
        else
        {
            if(v==id)
            {
                flag1=1;
                break;
            }
        }
    }
    if(!sum&&!flag&&!flag1)return 1;
    else return 0;
}
int main()
{

	scanf("%d%d", &m, &id);
	scanf("%s%s", b, c);

	memset(dp, -1, sizeof(dp));

	printf("%lld\n", (go(c)-go(b)+pan(b)+mod)%mod);

	return 0;
}