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 = 1000000007;
long long INV2 = 500000004;
long long INV6 = 166666668;
long long power(long long a, long long b, long long c) {
  long long x = 1, y = a;
  while (b > 0) {
    if (b & 1) x = (x * y) % MOD;
    y = (y * y) % MOD;
    b /= 2;
  }
  return x % c;
}
map<char, int> m, arr;
int main() {
  string s;
  cin >> s;
  int val = 0;
  long long ans = 1;
  for (char ch = '0'; ch <= '9'; ++ch) m[ch] = val++;
  for (char ch = 'A'; ch <= 'Z'; ch++) m[ch] = val++;
  for (char ch = 'a'; ch <= 'z'; ++ch) m[ch] = val++;
  m['-'] = 62;
  m['_'] = 63;
  char au[65];
  memset(au, '0', sizeof(au));
  int len = s.size();
  int idx = 0;
  for (int i = 0; i <= 63; ++i) {
    for (int j = 0; j <= 63; ++j) {
      arr[i & j]++;
    }
  }
  for (int i = len - 1; i >= 0; --i) {
    int val = m[s[i]];
    ans = (ans * arr[val]) % MOD;
  }
  printf("%lld\n", ans);
  return 0;
}
