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:
def intervalo(c,p,s):
    return ord(c) >= ord(p) and ord(c) <= ord(s)

s = input()
mod = 1000000007
saida = 1

for e in s:
    if(intervalo(e,'0','9')):
        e = ord(e) - ord('0')
    elif(intervalo(e,'a','z')):
        e = ord(e) - ord('a') + 36
    elif(intervalo(e,'A','Z')):
        e = ord(e) - ord('A') + 10
    elif(e == '-'):
        e = 62
    else:
        e = 63
    for i in range(6):
        if(((e >> i)&1) == 0):
            saida = (3*saida)%mod

print(saida)