Codeforces Round #265 (Div. 2)E（数位dp）

C. Substitutes in Number

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

Andrew and Eugene are playing a game. Initially, Andrew has string s, consisting of digits. Eugene sends Andrew multiple queries of type "di → ti",
that means "replace all digits di in
string s with substrings equal to ti".
For example, if s = 123123, then query "2 → 00" transforms s to 10031003,
and query "3 → " ("replace 3 by an empty string") transforms it to s = 1212.
After all the queries Eugene asks Andrew to find the remainder after division of number with decimal representation equal to s by 1000000007 (109 + 7).
When you represent s as a decimal number, please ignore the leading zeroes; also if s is
an empty string, then it's assumed that the number equals to zero.

Andrew got tired of processing Eugene's requests manually and he asked you to write a program for that. Help him!

Input

The first line contains string s (1 ≤ |s| ≤ 105),
consisting of digits — the string before processing all the requests.

The second line contains a single integer n (0 ≤ n ≤ 105) —
the number of queries.

The next n lines contain the descriptions of the queries. The i-th
query is described by string "di->ti",
where di is
exactly one digit (from 0 to 9), ti is
a string consisting of digits (ti can
be an empty string). The sum of lengths of ti for
all queries doesn't exceed 105. The
queries are written in the order in which they need to be performed.

Output

Print a single integer — remainder of division of the resulting number by 1000000007 (109 + 7).

Sample test(s)

input

123123
1
2->00

output

10031003

input

123123
1
3->

output

1212

input

222
2
2->0
0->7

output

777

input

1000000008
0

output

1

Note

Note that the leading zeroes are not removed from string s after the replacement (you can see it in the third sample).

题意：RT

思路：显然给出的每种转化关系前面的一定会被后面的覆盖，也就是说后面的信息才是准确的

那么也就是说可以将这些关系式倒着搞，然后分别维护0-9这些数字的pow值和val值

val代表这个数字的权重

pow代表这个数字的位数的值

比如2->123，初始pow[2]=10，val[2]=2

经过这个关系式转换后变成，pow[2]=1000，val[2]=123

处理完这些式子以后for一遍原串，用0-9这些数字的这两个值算出即可