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;
void bad(string mes = "Impossible") {
  cout << mes;
  exit(0);
}
template <typename... T>
void shit(T&... x) {}
template <typename... T>
void dec(T&... x) {
  shit(--x...);
}
template <typename... T>
void inc(T&... x) {
  shit(++x...);
}
template <typename T>
string bin(T x, int st = 2) {
  string ans = "";
  while (x > 0) {
    ans += char('0' + x % st);
    x /= st;
  }
  reverse(ans.begin(), ans.end());
  return ans.empty() ? "0" : ans;
}
template <typename T>
T dcm(string& second) {
  T x = 0;
  for (int i = 0; i < second.size(); i++) {
    x = (x * 2) + (second[i] == '1');
  }
  return x;
}
template <typename T>
T input() {
  T ans = 0, m = 1;
  char c = ' ';
  while (c == ' ' || c == '\n') c = getchar();
  if (c == '-') m = -1, c = getchar();
  while (c >= '0' && c <= '9') {
    ans = ans * 10 + int(c - '0'), c = getchar();
  }
  return ans * m;
}
const int inf = 1e9;
const double eps = 1e-9;
const int maxn = 3e2 + 10, base = 1e9 + 7;
const int sigm = 26;
const long long llinf = 1e18;
template <typename T>
T binpow(T n, T second) {
  if (second <= 0) return 1LL;
  if (second % 2 == 0) {
    T b = binpow(n, second / 2);
    return (1LL * b * b) % base;
  } else {
    return (1LL * binpow(n, second - 1) * n) % base;
  }
}
int prs(char c) {
  if (c >= '0' && c <= '9') return c - '0';
  if (c >= 'A' && c <= 'Z') return c - 'A' + 10;
  if (c >= 'a' && c <= 'z') return c - 'a' + 36;
  if (c == '-') return 62;
  if (c == '_') return 63;
}
int main() {
  string notParsedShit;
  cin >> notParsedShit;
  long long ans = 1;
  for (int i = 0; i < notParsedShit.size(); i++) {
    int cur = prs(notParsedShit[i]);
    for (int j = 0; j < 6; j++) {
      if (cur % 2 == 0) ans = (3LL * ans) % base;
      cur /= 2;
    }
  }
  cout << ans;
  return 0;
}