#include <bits/stdc++.h>
const int inf = 0x3f3f3f3f, maxn = 100000 + 5, mod = 1000000000 + 7;
using namespace std;
map<char, int> m;
char s[maxn];
int a[70];
int main() {
  memset(a, 0, sizeof(a));
  for (int i = 0; i <= 63; i++) {
    for (int j = 0; j <= 63; j++) {
      a[i & j]++;
    }
  }
  for (int i = 1; i <= 63; i++) {
    if (i < 10) {
      m['0' + i] = i;
    } else if (i >= 10 && i <= 35) {
      m[i - 10 + 'A'] = i;
    } else if (i >= 36 && i <= 61) {
      m[i - 36 + 'a'] = i;
    } else if (i == 62) {
      m['-'] = i;
    } else
      m['_'] = i;
  }
  while (cin >> s) {
    long long ans = 1;
    int len = strlen(s);
    for (int i = 0; i < len; i++) {
      ans = ans * a[m[s[i]]] % mod;
    }
    printf("%lld\n", 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>
const int inf = 0x3f3f3f3f, maxn = 100000 + 5, mod = 1000000000 + 7;
using namespace std;
map<char, int> m;
char s[maxn];
int a[70];
int main() {
  memset(a, 0, sizeof(a));
  for (int i = 0; i <= 63; i++) {
    for (int j = 0; j <= 63; j++) {
      a[i & j]++;
    }
  }
  for (int i = 1; i <= 63; i++) {
    if (i < 10) {
      m['0' + i] = i;
    } else if (i >= 10 && i <= 35) {
      m[i - 10 + 'A'] = i;
    } else if (i >= 36 && i <= 61) {
      m[i - 36 + 'a'] = i;
    } else if (i == 62) {
      m['-'] = i;
    } else
      m['_'] = i;
  }
  while (cin >> s) {
    long long ans = 1;
    int len = strlen(s);
    for (int i = 0; i < len; i++) {
      ans = ans * a[m[s[i]]] % mod;
    }
    printf("%lld\n", ans);
  }
  return 0;
}

```