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
#!/usr/bin/python

#
# Generic memoizer
#
def memoize(f):
    cache= {}
    def memf(*x):
        if x not in cache:
            cache[x] = f(*x)
        return cache[x]
    return memf

w = raw_input()

M = [None] * 128
for i, v in enumerate("0123456789"):
    M[ord(v)] = i

M[ord('-')] = 62
M[ord('_')] = 63

for i, v in enumerate("abcdefghijklmnopqrstuvwxyz"):
    M[ord(v)] = 36 + i

for i, v in enumerate("ABCDEFGHIJKLMNOPQRSTUVWXYZ"):
    M[ord(v)] = 10 + i

MOD = 10**9 + 7

@memoize
def nmatches(nzeros):
    c = 0
    for x in range(1<<nzeros):
        for y in range(1<<nzeros):
            if (x & y) != 0:
                continue
            c += 1
    return c

#for k in range(10):
#    print k, nmatches(k)

zbits = 1
for _w in w:
    v = M[ord(_w)]
    b = bin(v)[2:]
    b = '0' * (6-len(b)) + b
    # print "v", v, b
    z = b.count("0")
    mm = nmatches(z)
    # print z
    zbits = (zbits * mm) % MOD

print zbits

# 1: 0 -> 00 01 10  = 3
# 2: 00 -> 00-00, 01-10, 10-01, 11-00, 01-00, 10-00

