codeforces 355(Div 2)C. Vanya and Label(思维，预处理)

题目链接：

http://codeforces.com/problemset/problem/677/C

C. 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 base64 and got new word. Now Vanya thinks of some strings and wants to know the number of pairs of words of length|s| (length of s), such that their bitwise AND is equal tos. As this number can be large, output it modulo10⁹ + 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 from0 to 9;
• letters from 'A' to 'Z' correspond to integers from10 to 35;
• letters from 'a' to 'z' correspond to integers from36 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 strings 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:

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

题意：给定一种特定的64进制如下：

• digits from '0' to '9' correspond to integers from0 to 9;
• letters from 'A' to 'Z' correspond to integers from10 to 35;
• letters from 'a' to 'z' correspond to integers from36 to 61;
• letter '-' correspond to integer 62;
• letter '_' correspond to integer 63.

现给你一个字符串,每个字符对应64进制里面的特定的数，求有多少对相同长度字符串能够做按位与运算得到所给的字符串。

解题思路：
其实目的很明确，就是求出每个字符对应的数，可以由多少对字符所对应的数按位与得到 ，再将每个字符的可能结果乘起来就可以得出整个字符串的可能结果。
我是利用了map的性质，把字符作为map数组的下标去处理，感觉直观很多。

代码如下：

#include<iostream>#include<algorithm>#include<cstdio>#include<cstring>#include<set>#include<cmath>#include<vector>#include<map>using namespace std;const int maxn = 1e+5 + 10;char str[maxn];typedef long long ll;char ss[]= {"0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz-_"};map<char,int>ma;const int mod = 1e+9 + 7;int main(){    ma.clear();    scanf("%s",str);    for(int i = 0; ss[i]; i++)    {        for(int j = 0; ss[j]; j++)            ma[ss[i&j]]++;    }    ll ans = 1;    for(int i = 0 ; str[i] ; i ++)    {//        cout << ma[str[i]] <<endl;        ans = ans*ma[str[i]]%mod;    }    cout << ans <<endl;    return 0 ;}