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 CPP):

Solution:
#include <bits/stdc++.h>
const long long mod = 1000000007;
using namespace std;
void sc(int &x) {
  register int c = getchar();
  x = 0;
  int neg = 0;
  for (; ((c < 48 || c > 57) && c != '-'); c = getchar())
    ;
  if (c == '-') {
    neg = 1;
    c = getchar();
  }
  for (; c > 47 && c < 58; c = getchar()) {
    x = (x << 1) + (x << 3) + c - 48;
  }
  if (neg) x = -x;
}
long long power(long long a, long long b) {
  long long res = 1;
  while (b > 0) {
    if (b % 2 != 0) res = (res * a) % mod;
    a = (a * a) % mod;
    b /= 2;
  }
  return res;
}
long long gcd(long long a, long long b) {
  return ((b == 0) ? a : gcd(b, a % b));
}
int main() {
  int i, j, k, n;
  long long f[70] = {0}, ans = 1;
  char s[100002];
  cin >> (s + 1);
  for (i = 0; i < 64; i++) {
    k = i;
    for (j = 0; j < 6; j++) {
      if (k % 2 == 0) f[i]++;
      k /= 2;
    }
    f[i] = power(3, f[i]);
  }
  n = strlen(s + 1);
  for (i = 1; i < n + 1; i++) {
    if (s[i] >= '0' && s[i] <= '9')
      ans = (ans * f[s[i] - '0']) % mod;
    else if (s[i] >= 'a' && s[i] <= 'z')
      ans = (ans * f[s[i] - 'a' + 36]) % mod;
    else if (s[i] >= 'A' && s[i] <= 'Z')
      ans = (ans * f[s[i] - 'A' + 10]) % mod;
    else if (s[i] == '_')
      ans = (ans * f[63]) % mod;
    else if (s[i] == '-')
      ans = (ans * f[62]) % mod;
  }
  cout << ans << endl;
  return 0;
}