Problem: 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

Solution (in PYTHON3):

Solution:
from collections import *
from math import floor, log2, ceil


def fill():
    tem, ch = defaultdict(int, {str(i): i for i in range(10)}), ord('A')
    for i in range(10, 36):
        tem[chr(ch)] = i
        ch += 1

    ch = ord('a')
    for i in range(36, 62):
        tem[chr(ch)] = i
        ch += 1

    tem['-'], tem['_'] = 62, 63
    return tem


s, mem, ans = input(), fill(), 1
for i in s:
    for j in range(6):
        if mem[i] & (1 << j) == 0:
            ans = (ans * 3) % 1000000007
print(ans)