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.

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 '_'.

Print a single integer — the number of possible pairs of words, such that their bitwise AND is equal to string s modulo 109 + 7.

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 ;

 }