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
mod = 1000000007
resp = 1
mapa = {"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}

string = raw_input() 

for i in string:
  for j in range(6):
    if((mapa[i] >> j) & 1) == 0:
      resp = (3*resp) % mod

print resp
		 	   	 	 	  						 	 		  	 		