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
import sys
import bisect
# import heapq
from math import ceil,floor,factorial

def c(a,b):
    return int((factorial(a))/((factorial(b)*factorial(a-b))))

RI = lambda : [int(x) for x in sys.stdin.readline().split()]
ri = lambda : sys.stdin.readline().strip()
mod = 10**9+7
# for _ in range(int(ri())):

dic = {}
for i in range(0,10):
    dic[chr(ord('0')+i)] = i
for i in range(10,36):
    dic[chr(ord('A')+i-10)] = i
for i in range(36,62):
    dic[chr(ord('a')+i-36)] = i
dic['-'] = 62
dic['_'] = 63

st = ri()
ans = 1
for i in range(len(st)):
    num = dic[st[i]]
    for i in range(0,6):
        if num>>i & 1 == 0:
            ans= (ans*3)%mod
print(ans)