Vanya and LabelCrawling in process... Crawling failed Time Limit:1000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

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uDebug

Description

 

Input

 

Output

 

Sample Input

 

Sample Output

 

Hint

 

Description

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.

Sample Input

 

Input

z

Output

3

Input

V_V

Output

9

Input

Codeforces

Output

130653412

Sample Output

 

Hint

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
   /*有几个0，就有几个三*/#include 
                
                 #include 
                 
                  #include 
                  
                   #include 
                   
                    #define N 100010using namespace std;const int mod =1e9+7;/*void inti(){    for(int i=0;i<=63;i++)    {        int s=0;        while(i)        {            if(i%2==0) s++;            i/=2;        }        ans[i]=s;    }}*/int get(char c)  {      if(c>='0'&&c<='9')return c-'0';      if(c>='A'&&c<='Z')return c-'A'+10;      if(c>='a'&&c<='z')return c-'a'+36;      if(c=='-')return 62;      if(c=='_')return 63;  }  int main(){    //freopen("in.txt","r",stdin);    char ch[N];    long long s=1;    scanf("%s",&ch);    for(int i=0;ch[i];i++)    {        long long p=get(ch[i]);        for(int j=0;j<6;j++)              if(!((p>>j)&1))                  s=s*3%mod; //只有三种    }    printf("%lld\n",s);    return 0;}