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 main():
    def f1(c):
        x = ord(c)
        if ord('0') <= x <= ord('9'):
            res = int(c)
        elif ord('A') <= x <= ord('Z'):
            res = x - ord('A') + 10
        elif ord('a') <= x <= ord('z'):
            res = x - ord('a') + 36
        elif c == '-':
            res = 62
        elif c == '_':
            res = 63
        return res

    def f2(x):
        res = bin(x)[2:]
        res = '0' * (6 - len(res)) + res
        return res

    read = lambda: map(int, input().split())
    s = input()
    mod = 10 ** 9 + 7
    n = len(s)
    a = [f2(f1(i)) for i in s]
    t = ''.join(a)
    p = t.count('0')
    ans = 3 ** p % mod
    print(ans)
main()