Problem: 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

Solution (in PYTHON3):

Solution:
s=input()
ans=[0 for i in range(64)]
for i in range(len(ans)):
    temp=bin(i)

    ans[i]=temp.count('0')-1+(6-(len(temp)-2))

pri=10**9+7


count=1
for i in range(len(s)):
    if(s[i].isdigit()):
        r=bin(int(s[i]))
        r=r[2:]
        count=count*pow(3,ans[int(s[i])],pri)
        count%=pri
    elif(s[i]=='-'):
        count=count*pow(3,ans[62],pri)
        
        count%=pri
    elif(s[i]=='_'):
        count=count*pow(3,ans[63],pri)
   
        count%=pri
    elif(65<=ord(s[i])<=91):
        count=count*pow(3,ans[ord(s[i])-65+10],pri)
 
        count%=pri
    elif(97<=ord(s[i])<=122):
        count=count*pow(3,ans[ord(s[i])-97+36],pri)

        count%=pri
print(count%pri)