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;
char a[100010], aa;
int main() {
  long long ans, i, k, kk, lp, gs, j, tot, cnt, p[10], la, cf;
  scanf("%s", a);
  la = strlen(a);
  ans = 1LL;
  for (i = 0; i < la; i++) {
    aa = a[i];
    if (aa >= '0' && aa <= '9')
      k = aa - '0';
    else if (aa >= 'A' && aa <= 'Z')
      k = aa - 'A' + 10;
    else if (aa >= 'a' && aa <= 'z')
      k = aa - 'a' + 36;
    else if (aa == '-')
      k = 62;
    else
      k = 63;
    kk = k;
    memset(p, 0, sizeof(p));
    lp = 0;
    while (kk > 0) {
      p[++lp] = kk % 2;
      kk /= 2;
    }
    gs = 0;
    for (j = 1; j <= 6; j++)
      if (p[j] == 0) gs++;
    for (j = 1; j <= gs; j++) ans = (ans * 3LL) % 1000000007;
  }
  printf("%I64d", ans);
  return 0;
}
