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:
num = list(range(10))
bword = list(range(10,36))
sword = list(range(36,62))

bigconst = 65
smallconst = 97
ans = 1
s= input()
for i in range(len(s)):
    if s[i]>='0' and s[i] <= '9':
        k = int(s[i])
    elif s[i] >='A' and s[i] <='Z':
        k = bword[ord(s[i]) - bigconst]
    elif s[i] >='a' and s[i] <='z':
        k = sword[ord(s[i])-smallconst]
    elif s[i] == '-':
        k = 62
    else:
        k = 63

    k = bin(k)
    ans = (ans*((3**(k.count('0') -1 + ( 6 - (len(k) -2)))))%(10**9+7))%(10**9+7)
print(ans%(10**9+7))