Codeforces Round #355 (Div. 2) C. Vanya and Label （打表+字符串）

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 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

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

题解：
有一个字符串，这个字符串的每个字符分别代表着0-63之间的数字，然后问有多少对相同长度的字符串 & （按位与）运算之后，恰好等于这个字符串，又因为 每个字符是独立的，如果每个字符的方案数都知道，然后再全部乘起来就好了。。。。

AC代码：

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
LL a[1010]={0};
LL mod=1e9+7;
string s;
LL ans;
int main()
{
  cin>>s;
  //打表 
  for(int i=0;i<=63;i++)
  {
    for(int j=0;j<=63;j++)
    {
      a[i&j]++;
    }
  }
  ans=1;
  
  for(int i=0;i<s.size();i++)
  {
    int p;
    if(s[i]>='0'&&s[i]<='9') p=s[i]-'0';
    if(s[i]>='A'&&s[i]<='Z') p=s[i]-'A'+10;
    if(s[i]>='a'&&s[i]<='z') p=s[i]-'a'+36;
    if(s[i]=='-') p=62;
    if(s[i]=='_') p=63;
    ans=ans*a[p];
    ans%=mod;
  }
  cout<<ans;
  return 0;
}