Codeforces Round #355 (Div. 2) C 预处理

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 possible solutions:

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

题意：不同字符表示0~63   给你一个字符串判断 有多少对 长度相同的字符串的 &运算的结果 为所给的字符串

题解：模拟

1&1=1   0&1=0  1&0=0  0&0=0

打表预处理 0~63的满足情况的对数

例如31=（11111）2

共有三对    111111&011111

011111&111111

011111&011111

结果乘积并取模

 #include<iostream>
 #include<cstring>
 #include<cstdio>
 #include<queue>
 #include<stack>
 #include<cmath>
 #include<map>
 #define ll __int64
 #define pi acos(-1.0)
 #define mod 1000000007
 using namespace std;
 map<char,int> mp;
 ll a[];
 char b[];
 ll quickmod(ll a,ll b)
 {
     ll sum=;
     while(b)
     {
         if(b&)
             sum=(sum*a%mod);
         b>>=;
         a=(a*a%mod);
     }
     return sum;
 }
 void init ()
 {
     for(int i=;i<=;i++)
     {
         int n=i;
         int c=;
          while (n >)
       {
          if((n &) ==)
             ++c ;
          n >>= ;
       }
       a[i]=quickmod(,-c);
     }
     int jishu=;
     for(int i='';i<='';i++)
      {
       mp[i]=a[jishu];
       jishu++;
       }
     for(int i='A';i<='Z';i++)
     {
       mp[i]=a[jishu];
       jishu++;
     }
     for(int i='a';i<='z';i++)
       {
       mp[i]=a[jishu];
       jishu++;
       }
     mp['-']=a[];
     mp['_']=a[];
 }
 int main()
 {
     init();
     ll ans=;
     scanf("%s",b);
     int len=strlen(b);
     for(int i=;i<len;i++)
         ans=(ans%mod*mp[b[i]])%mod;
     cout<<ans<<endl;
     return ;
 }