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 = 1e18 + 1;
inline long long input(void) {
  char t;
  long long x = 0;
  int neg = 0;
  t = getchar();
  while ((t < 48 || t > 57) && t != '-') t = getchar();
  if (t == '-') {
    neg = 1;
    t = getchar();
  }
  while (t >= 48 && t <= 57) {
    x = (x << 3) + (x << 1) + t - 48;
    t = getchar();
  }
  if (neg) x = -x;
  return x;
}
inline void output(long long x) {
  char a[20];
  int i = 0, j;
  a[0] = '0';
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  if (x == 0) putchar('0');
  while (x) {
    a[i++] = x % 10 + 48;
    x /= 10;
  }
  for (j = i - 1; j >= 0; j--) {
    putchar(a[j]);
  }
  putchar('\n');
}
char s[1000010];
int main() {
  long long n, ans = 1, x;
  scanf("%s", s);
  n = strlen(s);
  for (int i = 0; i < n; ++i) {
    if (s[i] <= '9' && s[i] >= '0')
      x = s[i] - '0';
    else if (s[i] <= 'Z' && s[i] >= 'A')
      x = s[i] - 'A' + 10;
    else if (s[i] <= 'z' && s[i] >= 'a')
      x = s[i] - 'a' + 36;
    else if (s[i] == '-')
      x = 62;
    else
      x = 63;
    for (int j = 0; j < 6; ++j) {
      if (x % 2 == 0) ans *= 3;
      x /= 2;
    }
    ans %= 1000000007;
  }
  output(ans);
  return 0;
}