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
import sys


def count(i):
    c = 0
    for d in range(6):
        c += int(i % 2 == 0)
        i /= 2
    return c


s = sys.stdin.next().strip()

c = 0
for ch in s:
    i = None
    s1 = (ord('0'), ord('9'))
    s2 = (ord('A'), ord('Z'))
    s3 = (ord('a'), ord('z'))

    n = ord(ch)
    if (n >= s1[0]) and (n <= s1[1]):
        i = n - s1[0]
    if (n >= s2[0]) and (n <= s2[1]):
        i = n - s2[0] + 10
    if (n >= s3[0]) and (n <= s3[1]):
        i = n - s3[0] + 36
    if ch == '-':
        i = 62
    if ch == '_':
        i = 63
    c += count(i)

r = 1
m = 10 ** 9 + 7
i = 0
while i < c:
    r = (r * 3) % m
    i += 1

print(r)
