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;
int getZeros(int x) {
  string s;
  while (x) {
    s += (char)('0' + x % 2);
    x /= 2;
  }
  int count = 0;
  for (int i = 0; i < s.length(); i++) {
    if (s[i] == '0') count++;
  }
  return (count + (6 - s.length()));
}
void buildMap(map<char, int> &m) {
  int i;
  for (i = 0; i < 10; i++) m['0' + i] = getZeros(i);
  for (int j = 0; j < 26; j++, i++) m['A' + j] = getZeros(i);
  for (int j = 0; j < 26; j++, i++) m['a' + j] = getZeros(i);
  m['-'] = getZeros(62);
  m['_'] = getZeros(63);
}
long long int exp(int x) {
  vector<int> v;
  while (x) {
    v.push_back(x % 2);
    x /= 2;
  }
  long long int d = 1;
  for (int i = v.size() - 1; i >= 0; i--) {
    d = (d * d) % 1000000007;
    if (v[i]) d = (d * 3) % 1000000007;
  }
  return d;
}
int main(void) {
  map<char, int> m;
  buildMap(m);
  char s[100001];
  scanf(" %s", s);
  int z = 0;
  for (int i = 0; s[i] != '\0'; i++) z += m[s[i]];
  long long int ans = exp(z);
  cout << ans << "\n";
  return 0;
}
