[暴力枚举]Codeforces Vanya and Label

Vanya and Label

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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

Copy

z

output

Copy

3

input

Copy

V_V

output

Copy

9

input

Copy

Codeforces

output

Copy

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:

1. z&_ = 61&63 = 61 = z
2. _&z = 63&61 = 61 = z
3. z&z = 61&61 = 61 = z

题意:给你一个字符串s，把字符映射为一些数字，问有多少对和这个字符串长度相同的字符串逐位AND得到仍能得到字符串s

思路:看每位有多少种可能，再把他们相乘并取模，就是答案，注意到最多只有64个字符，所以O(n*n)暴力，4096*1e5大概在4e8可以在一秒内跑完

#include<bits/stdc++.h>
using namespace std;
const int amn=1e5+5,mod=1e9+7;
char a[amn];
int main(){
    ios::sync_with_stdio(0);
    cin>>a;
    long long ans=1,sum;
    int len=strlen(a),in,jg;
    for(int i=0;i<len;i++){
        if(a[i]>='0'&&a[i]<='9')in=a[i]-'0';
        else if(a[i]>='A'&&a[i]<='Z')in=a[i]-'A'+10;
        else if(a[i]>='a'&&a[i]<='z')in=a[i]-'a'+36;
        else if(a[i]=='-')in=62;
        else in=63;
        sum=0;
        for(int j=0;j<=63;j++){
            for(int k=0;k<=63;k++){
                jg=j&k;
                if(jg==in)sum++;
            }
        }
        ans=((ans%mod)*(sum%mod))%mod;
    }
    printf("%lld
",ans);
}
/***
给你一个字符串s，把字符映射为一些数字，问有多少对和这个字符串长度相同的字符串逐位AND得到仍能得到字符串s
看每位有多少种可能，再把他们相乘并取模，就是答案，注意到最多只有64个字符，所以O(n*n)暴力，4096*1e5大概在4e8可以在一秒内跑完
***/