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
#include <bits/stdc++.h>
using namespace std;
int dr[] = {1, 1, 0, -1, -1, -1, 0, 1};
int dc[] = {0, 1, 1, 1, 0, -1, -1, -1};
const double EPS = 1e-9;
const double Pi = acos(-1);
long long GCD(long long a, long long b) { return b == 0 ? a : GCD(b, a % b); }
long long LCM(long long a, long long b) { return a * (b / GCD(a, b)); }
int a[500];
long long BigMod(long long a, long long p) {
  if (!p) return 1;
  if (p & 1) return (BigMod(a, p - 1) % 1000000007 * a) % 1000000007;
  long long x = BigMod(a, p >> 1) % 1000000007;
  return (x * x) % 1000000007;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  string s;
  cin >> s;
  a['-'] = 62;
  a['_'] = 63;
  for (char i = '0'; i <= '9'; i++) a[i] = i - '0';
  for (char i = 'A', j = 10; i <= 'Z'; i++) a[i] = j++;
  for (char i = 'a', j = 36; i <= 'z'; i++) a[i] = j++;
  long long zeros = 0;
  for (int i = 0; i < (int)s.size(); i++)
    for (int b = 5; b >= 0; b--)
      if (!(a[s[i]] & (1 << b))) zeros++;
  cout << BigMod(3, zeros) << endl;
  return 0;
}
