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>
using namespace std;
const long long INF = 1000000000 + 7;
const double esp = 1e-13;
long long nhan(long long x, long long y, long long m) {
  long long ans = 0;
  while (y) {
    if (y % 2) ans = (ans + x) % m;
    x = x * 2 % m;
    y /= 2;
  }
  return ans;
}
long long power(long long a, long long n, long long m) {
  unsigned long long ans = 1;
  while (n) {
    if (n % 2) ans = ans * a % m;
    a = a * a % m;
    n /= 2;
  }
  return ans;
}
bool check(long long n) {
  if (n == 2) return true;
  if (n < 2 || n % 2 == 0) return false;
  long long p[3] = {3, 5, 7};
  long long a, d;
  int r, s;
  d = n - 1;
  s = 0;
  while (d % 2 == 0) {
    d /= 2;
    s++;
  }
  for (long long i = 0; i <= 2; i++) {
    if (n == p[i]) return true;
    if (n % p[i] == 0) return false;
    a = power(p[i], d, n);
    if (a != 1) {
      for (r = 0; r < s && a != n - 1; r++) a = a * a % n;
      if (r == s) return false;
    }
  }
  return true;
}
long long phi(long long n) {
  long long ans = n;
  long long n2 = (long long)sqrt(n);
  for (long long i = 2; i <= n2; i++)
    if (n % i == 0) {
      while (n % i == 0) n /= i;
      ans -= ans / i;
    }
  if (n > 1) ans -= ans / n;
  return ans;
}
int get(char c) {
  if ('0' <= c && c <= '9') return c - '0';
  if ('A' <= c && c <= 'Z') return c - 'A' + 10;
  if ('a' <= c && c <= 'z') return c - 'a' + 36;
  if (c == '-') return 62;
  if (c == '_') return 63;
}
long long ans, f[1000];
char s[100000 + 10];
int n;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie();
  srand(time(0));
  ans = 1;
  cin >> s;
  for (long long i = 0; i <= 63; i++)
    for (long long j = 0; j <= 63; j++) f[i & j]++;
  n = strlen(s);
  for (long long i = 0; i < n; i++) ans = ans * f[get(s[i])] % INF;
  cout << ans;
}