Shash and Exun Celebrations

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Shash is the richest man in the world by a long shot. Exun 2042, DPS RK Puram’s 42nd (assume it was cancelled deliberately for 5 years to sync up ;) ) Tech Festival is coming and Shash is delighted. He decides to pay his employees extra to celebrate the occasion. So he decides to draw a cheque with a value V that is to be distributed amongst his employees.
 V must be an integral value between A and B (inclusive). The number of digits in A and B is equal.
 Suppose Shash assigns 1 based indexing to the digits of this number V. He doesn’t like digits at indexes divisible by K. Neither does he like the digit L. So he puts L only at indexes divisible by K, and nowhere else.
 Shash Evil Incorporative has M employees and 1 teaboy. He must give an equal integral amount to all the M employees and the teaboy must be paid a fixed value X. X will always be less than M. The whole value V should be used up in making this payment.
Find the number of V’s (mod 1e9+7) that Shash can choose such that these constraints are satisfied.
Input
The first line contains 4 integers, K, L, M and X as described above.
The second line contains the lower limit on V in the form of a string A.
The third line contains the upper limit on V in the form of a string B.
Output
A single integer, the number of possible values mod 1e9+7
Constraints
 1 <= A = B <= 1,000
 K>1
 X < M <= 1,000
 0<=L<=9
A, B <= 8
Subtask 2 (20 Points):
M = 2, 5 or 10
Subtask 3 (70 Points)
Original Constraints
Example: Sample Input 1: 2 5 6 2 100 199 Sample Output 1: 2 Explanation 1: Only 2 numbers, 152 and 158 satisfy the constraints. For 152, 30 are given to each employee. For 158, 31 are given to each employee. Both have the digit 5 only in places divisible by 2. Sample Input 2: 3 5 5 0 100 200 Sample Output 2: 9 Explanation 2: 105, 115, 125, 135, 145, 165, 175, 185, 195 are the possible values. Notice that 155 also has 5 in the 2nd place and 2 is not divisible by 3, so it is not counted. Sample Input 3: 2 5 2 1 1 9 Sample Output 3: 4 Explanation 3: 1 3 7 9 are the possible values. Again notice that 5 cannot be a possible value as it can only be in positions divisible by 2.
Author:  harrycoda 
Tags  harrycoda 
Date Added:  16112017 
Time Limit:  1 sec 
Source Limit:  50000 Bytes 
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