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
s = raw_input()

l = len(s)

m=1000000007

def get(x):
	k = ord(x)
	if (k>=ord('0') and k<=ord('9')):
		return k-48
	elif (k>=ord('A') and k<=ord('Z')):
		return k-55
	elif (k>=ord('a') and k<=ord('z')):
		return k-61
	elif (x=='-'):
		return 62
	else:
		return 63

def exp(a,b):
	if b==0:
		return 1
	if (b%2==0):
		p = exp(a,b/2)
		return (p*p)%m
	return (a*exp(a,b-1))%m

k=0

for i in range(l):
	num = get(s[i])
	#print num
	t=0
	zz=0
	while(t<6):
		if num%2==0:
			zz+=1
		num/=2
		t+=1
	k+=zz
	#print zz

res = exp(3,k)

print res



