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
z = input()
l = []
for x in z:
	if(ord(x)>=ord('0') and ord(x)<=ord('9')):
		l.append(ord(x)-ord('0'))
	elif(ord(x)>=ord('A') and ord(x)<=ord('Z')):
		l.append(ord(x)-ord('A')+10)
	elif(ord(x)>=ord('a') and ord(x)<=ord('z')):
		l.append(ord(x)-ord('a')+36)
	elif(ord(x)==ord('-')):
		l.append(62)
	else:
		l.append(63)


ans=1
for i in l:
	unset = 6-bin(i).count('1')
	for x in range(unset):
		ans *= 3
		while(ans>=1000000007):
			ans-=1000000007
		# ans%=1000000007
print(ans)