Description
While walking down the street Vanya saw a label "Hide&Seek". Because he is a programmer, he used & as a bitwise AND for these two words represented as a integers in base 64 and got new word. Now Vanya thinks of some string s and wants to know the number of pairs of words of length |s| (length of s), such that their bitwise AND is equal to s. As this number can be large, output it modulo 109 + 7.
To represent the string as a number in numeral system with base 64 Vanya uses the following rules:
- digits from ‘0‘ to ‘9‘ correspond to integers from 0 to 9;
- letters from ‘A‘ to ‘Z‘ correspond to integers from 10 to 35;
- letters from ‘a‘ to ‘z‘ correspond to integers from 36 to 61;
- letter ‘-‘ correspond to integer 62;
- letter ‘_‘ correspond to integer 63.
Input
The only line of the input contains a single word s (1 ≤ |s| ≤ 100 000), consisting of digits, lowercase and uppercase English letters, characters ‘-‘ and ‘_‘.
Output
Print a single integer — the number of possible pairs of words, such that their bitwise AND is equal to string s modulo 109 + 7.
Examples
input
z
output
3
input
V_V
output
9
input
Codeforces
output
130653412
Note
For a detailed definition of bitwise AND we recommend to take a look in the corresponding article in Wikipedia.
In the first sample, there are 3 possible solutions:
- z&_ = 61&63 = 61 = z
- _&z = 63&61 = 61 = z
- z&z = 61&61 = 61 = z
给我们一下字符串，问我们可以通过&后，值依然和题目上规定的相同
我们当然可以先把0~63的先全部&一下，存下结果，然后，就一个字符一个字符的判断就好啦~
1 #include<iostream>
2 #include<stdio.h>
3 #include<map>
4 using namespace std;
5 map<int,int> q;
6 int main()
7 {
8 for(int i=0;i<=63;i++)
9 {
10 for(int j=0;j<=63;j++)
11 {
12 q[i&j]++;
13 }
14 }
15 int pos;
16 long long num=1;
17 string s;
18 cin>>s;
19 for(int i=0;i<s.length();i++)
20 {
21 if(s[i]>=‘0‘&&s[i]<=‘9‘)
22 {
23 pos=s[i]-‘0‘;
24 }
25 else if(s[i]>=‘A‘&&s[i]<=‘Z‘)
26 {
27 pos=s[i]-‘A‘+10;
28 }
29 else if(s[i]>=‘a‘&&s[i]<=‘z‘)
30 {
31 pos=s[i]-‘a‘+36;
32 }
33 else if(s[i]==‘-‘)
34 {
35 pos=62;
36 }
37 else if(s[i]==‘_‘)
38 {
39 pos=63;
40 }
41 num=num*q[pos]%1000000007;
42 }
43 cout<<num<<endl;
44 return 0;
45 }