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:
A = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz-_"
Z = [0] * 64
for i in range(64):
    r = 0
    for j in range(64):
        for k in range(64):
            if i == (j & k):
                r += 1
    Z[i] = r
    #print(str(i) + ".", A[i], "=>", r)

r = 1
for c in input():
    #print("c:", c, "idx:", A.index(c), "z:", Z[A.index(c)])
    r = (r * Z[A.index(c)]) % 1000000007
print(r)

"""
V_V & V_V = V_V

__V & V_V = V_V
V_V & __V = V_V

V__ & V_V = V_V
V_V & V__ = V_V

___ & V_V = V_V
V_V & ___ = V_V

__V & V__ = V_V
V__ & __V = V_V
"""