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 max_n = 111111, mod = 1000000007;
int n, c[11][11];
char s[max_n];
bool get_bit(int x, int poz) { return (bool)(x & (1 << poz)); }
int get(char c) {
  if ('0' <= c && c <= '9') {
    return c - '0';
  }
  if ('A' <= c && c <= 'Z') {
    return c - 'A' + 10;
  }
  if ('a' <= c && c <= 'z') {
    return c - 'a' + 36;
  }
  if (c == '-') {
    return 62;
  }
  return 63;
}
void get_all_c() {
  c[0][0] = 1;
  for (int i = 1; i < 11; ++i) {
    c[i][0] = 1;
    for (int j = 1; j <= i; ++j) {
      c[i][j] = (c[i - 1][j - 1] + c[i - 1][j]) % mod;
    }
  }
}
int main() {
  get_all_c();
  scanf("%s", s);
  n = strlen(s);
  int ans = 1;
  for (int i = 0; i < n; ++i) {
    int x = get(s[i]), cnt = 6;
    for (int j = 0; j < 6; ++j) {
      cnt -= get_bit(x, j);
    }
    int res = 0;
    for (int j = 0; j <= cnt; ++j) {
      res += ((1LL << (cnt - j)) * c[cnt][j]) % mod;
      res %= mod;
    }
    ans = (1LL * ans * res) % mod;
  }
  printf("%d\n", ans);
  return 0;
}