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
from math import factorial

b = [0, 1, 1, 2, 1, 2, 2, 3, 1, 2, 2, 3, 2, 3, 3, 4, 1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5, 1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5, 2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6]

d = {'N': 23, '9': 9, 'g': 42, '3': 3, '4': 4, 'Q': 26, 'K': 20, 'S': 28, 'A': 10, 'x': 59, 'B': 11, 'I': 18, 'n': 49, '7': 7, '-': 62, 'J': 19, 'm': 48, '5': 5, 'O': 24, 'L': 21, 'u': 56, 'r': 53, 'Z': 35, 'M': 22, 'j': 45, 'v': 57, 'q': 52, 'X': 33, 't': 55, 'a': 36, 's': 54, 'z': 61, 'Y': 34, 'd': 39, 'p': 51, 'o': 50, 'f': 41, 'P': 25, 'V': 31, '_': 63, 'w': 58, 'C': 12, 'R': 27, 'b': 37, 'E': 14, '0': 0, 'G': 16, 'k': 46, 'l': 47, 'D': 13, 'W': 32, 'T': 29, '8': 8, 'U': 30, 'F': 15, '2': 2, 'c': 38, 'h': 43, 'y': 60, 'i': 44, 'e': 40, '6': 6, 'H': 17, '1': 1}

MOD = 1e9 + 7
max_dig = 6
ans = 1

s = input()
for c in s:
    nul_num = max_dig - b[d[c]]
    ans *= 3**nul_num
    ans %= MOD
    
print(int(ans))
