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 int mod = 1e9 + 7;
int cnt;
int qucik_mod(int a, int b) {
  int res = 1;
  while (b) {
    if (b & 1) res = (long long)(res * a) % mod;
    b >>= 1;
    a = ((long long)a * a) % mod;
  }
  return res;
}
char s[212345];
int sum[212345];
int zhuanhuan(char c) {
  if (isdigit(c)) {
    return c - '0';
  } else if (isupper(c)) {
    return 10 + c - 'A';
  } else if (islower(c)) {
    return 36 + c - 'a';
  } else if (c == '_')
    return 63;
  else
    return 62;
}
int main() {
  for (int i = 0; i < (1 << 6); i++) {
    for (int j = 0; j < (1 << 6); j++) {
      sum[i & j] += 1;
    }
  }
  scanf("%s", s);
  int ans = 1;
  for (int i = 0; s[i]; i++) {
    int tem = zhuanhuan(s[i]);
    ans = ((long long)ans * sum[tem]) % mod;
  }
  cout << ans << '\n';
  return 0;
}