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;
const long long MOD = 1e9 + 7;
int HASH[1000];
long long Num[70];
void INIT() {
  int cnt = 0;
  for (int i = 0; i <= 9; ++i) HASH[i + '0'] = cnt++;
  for (int i = 0; i < 26; ++i) HASH[i + 'A'] = cnt++;
  for (int i = 0; i < 26; ++i) HASH[i + 'a'] = cnt++;
  HASH['-'] = cnt++;
  HASH['_'] = cnt;
  for (int i = 0; i <= cnt; ++i) {
    for (int j = i + 1; j <= cnt; ++j) {
      Num[i & j] += 2;
    }
  }
  for (int i = 0; i <= 63; ++i) Num[i]++;
}
int main() {
  INIT();
  string Str;
  cin >> Str;
  long long ANS = 1;
  int len = Str.length();
  for (int i = 0; i < len; ++i) {
    ANS = ANS * Num[HASH[Str[i]]] % MOD;
  }
  cout << ANS << endl;
}
