codeforces 677C Vanya and Label

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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

把所给的字符串的每个字符翻译成二进制的形式，让你求出相同的字符串长度下，与原字符串做&运算，使得最后结果还是为原来的字符串

1&0=1,0&1=1,0&0=0

所以关键就是求字符串翻译成二进制出现了多少个0

需要注意的是每个字符都要翻译成6位二进制（我就出错了一次，后来才发现的）

还有那就是注意整数快速幂

#include<iostream>#include<cstdio>#include<cstring>#include<cstdlib>#include<cmath>#include<string>#include<vector>#include<stack>#include<set>#include<map>#include<queue>#include<algorithm>using namespace std;char s[100005];int main(){int len;int i,j;int ans;scanf("%s",s);len=strlen(s);char c;int num;ans=0;for(i=0;i<len;i++){c=s[i];//printf("%c",c);if(c>='0'&&c<='9'){num=c-'0';}else if(c>='A'&&c<='Z'){num=c-'A'+10;}else if(c>='a'&&c<='z'){num=c-'a'+36;}else if(c=='-'){num=62;}else{num=63;}//printf("%d ",num);int t=6;while(t--){if(num%2==0){ans++;}num>>=1;}//printf("%d\n",ans);}long long re=1;long long a=3;//printf("%d\n",ans);while(ans!=0){if(ans%2==1){re=re*a%1000000007;}ans/=2;a=a*a%1000000007;}printf("%lld\n",re);    return 0;}

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