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>
const int maxn = 1e5 + 10;
const int dis[4][2] = {{0, 1}, {-1, 0}, {0, -1}, {1, 0}};
using namespace std;
long long gcd(long long p, long long q) { return q == 0 ? p : gcd(q, p % q); }
long long qpow(long long p, long long q, long long mo) {
  long long f = 1;
  while (q) {
    if (q & 1) f = f * p % mo;
    p = p * p % mo;
    q >>= 1;
  }
  return f;
}
char a[maxn];
int f[64];
void init() {
  for (int i = 0; i <= 63; i++) {
    int j = i;
    int cnt = 0;
    for (int k = 0; k < 6; k++) {
      if (!(j & 1)) cnt++;
      j >>= 1;
    }
    f[i] = cnt;
  }
}
int main() {
  int i, j, m, n, k;
  init();
  scanf("%s", a);
  long long p = 0;
  for (i = 0; a[i]; i++) {
    if (a[i] >= '0' && a[i] <= '9')
      p += f[a[i] - '0'];
    else if (a[i] >= 'a' && a[i] <= 'z')
      p += f[a[i] - 'a' + 36];
    else if (a[i] >= 'A' && a[i] <= 'Z')
      p += f[a[i] - 'A' + 10];
    else if (a[i] == '-')
      p += f[62];
    else
      p += f[63];
  }
  printf("%lld\n", qpow(1LL * 3, p, 1000000007));
  return 0;
}
