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


''' We can see that each 1 bit only gives us 1 pair (1,1) while a 0 gives us
3 pairs (1,0), (0,1) (0,0)  so each 0 in the representation can give us 3 different pairs

Each of the pairs of lets say an earlier 0 can be combined with each pair or a subsequent 0, so the total
amount becomes 3**(number of zeros)'''

amount_of_zeros = sum((6 - bin(numRep[c]).count("1") for c in raw_input().strip()))

print(3**amount_of_zeros % (10**9 + 7))
