#include <bits/stdc++.h>
using namespace std;
char a[100005];
long long exp(int a) {
  long long l = 1;
  for (int i = 1; i <= a; i++) {
    l = (l * 3) % 1000000007;
  }
  return l;
}
int main() {
  int i, j, k, len, sum = 0;
  map<char, int> q;
  q['0'] = 6;
  q['1'] = 5;
  q['2'] = 5;
  q['3'] = 4;
  q['4'] = 5;
  q['5'] = 4;
  q['6'] = 4;
  q['7'] = 3;
  q['8'] = 5;
  q['9'] = 4;
  q['A'] = 4;
  q['B'] = 3;
  q['C'] = 4;
  q['D'] = 3;
  q['E'] = 3;
  q['F'] = 2;
  q['G'] = 5;
  q['H'] = 4;
  q['I'] = 4;
  q['J'] = 3;
  q['K'] = 4;
  q['L'] = 3;
  q['M'] = 3;
  q['N'] = 2;
  q['O'] = 4;
  q['P'] = 3;
  q['Q'] = 3;
  q['R'] = 2;
  q['S'] = 3;
  q['T'] = 2;
  q['U'] = 2;
  q['V'] = 1;
  q['W'] = 5;
  q['X'] = 4;
  q['Y'] = 4;
  q['Z'] = 3;
  q['a'] = 4;
  q['b'] = 3;
  q['c'] = 3;
  q['d'] = 2;
  q['e'] = 4;
  q['f'] = 3;
  q['g'] = 3;
  q['h'] = 2;
  q['i'] = 3;
  q['j'] = 2;
  q['k'] = 2;
  q['l'] = 1;
  q['m'] = 4;
  q['n'] = 3;
  q['o'] = 3;
  q['p'] = 2;
  q['q'] = 3;
  q['r'] = 2;
  q['s'] = 2;
  q['t'] = 1;
  q['u'] = 3;
  q['v'] = 2;
  q['w'] = 2;
  q['x'] = 1;
  q['y'] = 2;
  q['z'] = 1;
  q['-'] = 1;
  q['_'] = 0;
  cin >> a;
  len = strlen(a);
  for (i = 0; i < len; i++) {
    sum += q[a[i]];
  }
  printf("%I64d\n", exp(sum));
  return 0;
}

### Prompt

Please formulate a CPP solution to the following 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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
char a[100005];
long long exp(int a) {
  long long l = 1;
  for (int i = 1; i <= a; i++) {
    l = (l * 3) % 1000000007;
  }
  return l;
}
int main() {
  int i, j, k, len, sum = 0;
  map<char, int> q;
  q['0'] = 6;
  q['1'] = 5;
  q['2'] = 5;
  q['3'] = 4;
  q['4'] = 5;
  q['5'] = 4;
  q['6'] = 4;
  q['7'] = 3;
  q['8'] = 5;
  q['9'] = 4;
  q['A'] = 4;
  q['B'] = 3;
  q['C'] = 4;
  q['D'] = 3;
  q['E'] = 3;
  q['F'] = 2;
  q['G'] = 5;
  q['H'] = 4;
  q['I'] = 4;
  q['J'] = 3;
  q['K'] = 4;
  q['L'] = 3;
  q['M'] = 3;
  q['N'] = 2;
  q['O'] = 4;
  q['P'] = 3;
  q['Q'] = 3;
  q['R'] = 2;
  q['S'] = 3;
  q['T'] = 2;
  q['U'] = 2;
  q['V'] = 1;
  q['W'] = 5;
  q['X'] = 4;
  q['Y'] = 4;
  q['Z'] = 3;
  q['a'] = 4;
  q['b'] = 3;
  q['c'] = 3;
  q['d'] = 2;
  q['e'] = 4;
  q['f'] = 3;
  q['g'] = 3;
  q['h'] = 2;
  q['i'] = 3;
  q['j'] = 2;
  q['k'] = 2;
  q['l'] = 1;
  q['m'] = 4;
  q['n'] = 3;
  q['o'] = 3;
  q['p'] = 2;
  q['q'] = 3;
  q['r'] = 2;
  q['s'] = 2;
  q['t'] = 1;
  q['u'] = 3;
  q['v'] = 2;
  q['w'] = 2;
  q['x'] = 1;
  q['y'] = 2;
  q['z'] = 1;
  q['-'] = 1;
  q['_'] = 0;
  cin >> a;
  len = strlen(a);
  for (i = 0; i < len; i++) {
    sum += q[a[i]];
  }
  printf("%I64d\n", exp(sum));
  return 0;
}

```