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  log
def f(s):
	s=s[::-1]
	a={}
	cons=0
	for i in range(10):
		a[str(i)]=cons
		cons+=1
	for i in "abcdefghijklmnopqrstuvwxyz".upper():
		a[i]=cons
		cons+=1
	for i in "abcdefghijklmnopqrstuvwxyz":
		a[i]=cons
		cons+=1
	# print(a)
	a["-"]=62
	a["_"]=63
	ans=1
	for i in s:
		pp=a[i]
		tc=bin(a[i]).lstrip("0b")
		tc="0"*(6-len(tc))+tc
		# print(tc,pp)
		ans*=(3**(tc.count("0")))
		ans%=(10**9+7)
	return ans

print(f(input()))