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;
inline int read() {
  int _x;
  scanf("%d", &_x);
  return _x;
}
const int INF = (int)1e9;
const int mod = (int)1e9 + 7;
const int maxn = (int)1e5 + 5;
const int logn = 20;
const int sqrtn = 1e3 + 5;
char q[maxn];
int N, ans[100];
int get(char a) {
  if (a >= '0' && a <= '9') return a - '0';
  if (a >= 'A' && a <= 'Z') return a - 'A' + 10;
  if (a >= 'a' && a <= 'z') return a - 'a' + 36;
  if (a == '-') return 62;
  return 63;
}
int mul(int a, int b) {
  long long int x = (long long int)a * b % mod;
  x %= mod;
  return x;
}
int main() {
  scanf("%s", q + 1);
  N = strlen(q + 1);
  for (int i = 0; i <= 63; i++)
    for (int j = 0; j <= 63; j++) ans[(i & j)]++;
  int res = 1;
  for (int i = 1; i <= N; i++) {
    int a = get(q[i]);
    res = mul(res, ans[a]);
  }
  cout << res << endl;
  return 0;
}
