codeforces628D. Magic Numbers (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-magic and 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

Note

The numbers from the answer of the first example are 16, 26, 36, 46, 56, 76, 86 and 96.

The numbers from the answer of the second example are 2, 4, 6 and 8.

The numbers from the answer of the third example are 1767, 2717, 5757, 6707, 8797 and 9747.

題意：給你一個區間[a,b]，讓你找到這個區間內滿足沒有前導零且偶數位都是d，奇數位不出現d，並且這個數能被m整除的數的個數。

思路：用dp[pos][yushu][oushu]表示pos位前面的位形成的數modm後餘數爲yushu，且當前位是否是偶數的方案數，要注意前導零。

#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<string>
#include<bitset>
#include<algorithm>
using namespace std;
typedef long long ll;
#define inf 99999999
#define MOD 1000000007
char s1[2005],s2[2005];
int wei[2005];
ll dp[2005][2005][2];
int m,d;
void add(ll& x,ll y) {
    x += y;
    if(x>=MOD) x-=MOD;
}


ll dfs(int pos,int yushu,int oushu,int flag,int zero)
{
    int i,j;
    if(pos==-1){
        if(zero==1)return 0;
        if(yushu==0)return 1;
        else return 0;
    }
    if(!flag && !zero && dp[pos][yushu][oushu]!=-1){
        return dp[pos][yushu][oushu];
    }

    int ed=flag?wei[pos]:9;
    ll ans=0;
    if(zero==1){
        add(ans,dfs(pos-1,yushu,oushu,0,1));
        for(i=1;i<=ed;i++){
            if(i!=d)add(ans,dfs(pos-1,(yushu*10+i)%m,1^oushu,flag&&wei[pos]==i,0) );
        }
    }
    else{
        if(oushu){
            if(d<=ed)add(ans,dfs(pos-1,(yushu*10+d)%m,1^oushu,flag&&wei[pos]==d,0) );
        }
        else{
            for(i=0;i<=ed;i++){
                if(i!=d)add(ans,dfs(pos-1,(yushu*10+i)%m,1^oushu,flag&&wei[pos]==i,0) );
            }
        }
    }
    if(!flag && !zero){
        dp[pos][yushu][oushu]=ans;
    }
    return ans;
}

ll solve(char s[])
{
    int len,i,j;
    len=strlen(s);
    for(i=len-1;i>=0;i--){
        wei[i]=s[i]-'0';
    }
    return dfs(len-1,0,0,1,1);
}

int main()
{
    int n,i,j,len1,len2;
    while(scanf("%d%d",&m,&d)!=EOF)
    {
        scanf("%s%s",s1,s2);
        len1=strlen(s1);
        reverse(s1,s1+len1);
        for(i=0;i<len1;i++){
            if(s1[i]=='0'){
                s1[i]='9';
            }
            else{
                s1[i]--;break;
            }
        }
        if(s1[len1-1]=='0'){
            s1[len1-1]='\0';
            len1--;
        }

        len2=strlen(s2);
        reverse(s2,s2+len2);
        memset(dp,-1,sizeof(dp));
        ll num1=solve(s1);
        ll num2=solve(s2);
        printf("%I64d\n",((num2-num1)%MOD+MOD)%MOD );

    }
    return 0;
}