#include <bits/stdc++.h>
using namespace std;
long long power(long long x, long long y) {
  if (y == 0)
    return 1;
  else if (y % 2 == 0)
    return power(x, y / 2) * power(x, y / 2);
  else
    return x * power(x, y / 2) * power(x, y / 2);
}
long long bigpower(long long x, long long y, long long p) {
  long long res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
long long gcd(long long a, long long b) {
  if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
long long at[2003];
int main() {
  char ar[100004];
  long long at[500];
  long long p = '0';
  for (long long i = 0; i <= 9; i++) at[p++] = i;
  p = 'A';
  for (long long i = 10; i <= 35; i++) at[p++] = i;
  p = 'a';
  for (long long i = 36; i <= 61; i++) at[p++] = i;
  at['-'] = 62;
  at['_'] = 63;
  cin >> ar;
  long long i;
  long long n = strlen(ar);
  long long ans = 1;
  long long count = 0;
  for (long long j = 0; j < n; j++) {
    count = 0;
    for (i = 0; i < 64; i++) {
      for (long long k = 0; k < 64; k++) {
        long long g = i & k;
        if (at[ar[j]] == g) count++;
      }
    }
    if (count > 0)
      ans = ((ans % 1000000007) * (count % 1000000007)) % 1000000007;
  }
  cout << ans;
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
long long power(long long x, long long y) {
  if (y == 0)
    return 1;
  else if (y % 2 == 0)
    return power(x, y / 2) * power(x, y / 2);
  else
    return x * power(x, y / 2) * power(x, y / 2);
}
long long bigpower(long long x, long long y, long long p) {
  long long res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
long long gcd(long long a, long long b) {
  if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
long long at[2003];
int main() {
  char ar[100004];
  long long at[500];
  long long p = '0';
  for (long long i = 0; i <= 9; i++) at[p++] = i;
  p = 'A';
  for (long long i = 10; i <= 35; i++) at[p++] = i;
  p = 'a';
  for (long long i = 36; i <= 61; i++) at[p++] = i;
  at['-'] = 62;
  at['_'] = 63;
  cin >> ar;
  long long i;
  long long n = strlen(ar);
  long long ans = 1;
  long long count = 0;
  for (long long j = 0; j < n; j++) {
    count = 0;
    for (i = 0; i < 64; i++) {
      for (long long k = 0; k < 64; k++) {
        long long g = i & k;
        if (at[ar[j]] == g) count++;
      }
    }
    if (count > 0)
      ans = ((ans % 1000000007) * (count % 1000000007)) % 1000000007;
  }
  cout << ans;
  return 0;
}

```