#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:102400000000,102400000000")
const long long MOD = 1000000007;
const int N = 1E5 + 15;
const int maxn = 8e3 + 15;
const int letter = 130;
const int INF = 2e9;
const double pi = acos(-1.0);
const double eps = 1e-10;
using namespace std;
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
int n, a[N], b[100];
long long bin[10];
int lowbit(int x) { return x & (-x); }
char s[N];
int main() {
  bin[0] = 1;
  for (int i = 1; i <= 8; i++) bin[i] = bin[i - 1] * 3;
  scanf("%s", s);
  n = strlen(s);
  for (int i = 0; i < n; i++) {
    if ('0' <= s[i] && s[i] <= '9')
      a[i] = s[i] - '0';
    else if ('A' <= s[i] && s[i] <= 'Z')
      a[i] = s[i] - 'A' + 10;
    else if ('a' <= s[i] && s[i] <= 'z')
      a[i] = s[i] - 'a' + 36;
    else if (s[i] == '-')
      a[i] = 62;
    else if (s[i] == '_')
      a[i] = 63;
  }
  long long sum = 1;
  for (int i = 1; i <= 63; i++) {
    int x = i;
    while (x) {
      b[i]++, x -= lowbit(x);
    }
  }
  for (int i = 0; i < n; i++) {
    sum = (sum * (1ll * bin[6 - b[a[i]]] % MOD)) % MOD;
  }
  printf("%I64d\n", sum % MOD);
  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>
#pragma comment(linker, "/STACK:102400000000,102400000000")
const long long MOD = 1000000007;
const int N = 1E5 + 15;
const int maxn = 8e3 + 15;
const int letter = 130;
const int INF = 2e9;
const double pi = acos(-1.0);
const double eps = 1e-10;
using namespace std;
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
int n, a[N], b[100];
long long bin[10];
int lowbit(int x) { return x & (-x); }
char s[N];
int main() {
  bin[0] = 1;
  for (int i = 1; i <= 8; i++) bin[i] = bin[i - 1] * 3;
  scanf("%s", s);
  n = strlen(s);
  for (int i = 0; i < n; i++) {
    if ('0' <= s[i] && s[i] <= '9')
      a[i] = s[i] - '0';
    else if ('A' <= s[i] && s[i] <= 'Z')
      a[i] = s[i] - 'A' + 10;
    else if ('a' <= s[i] && s[i] <= 'z')
      a[i] = s[i] - 'a' + 36;
    else if (s[i] == '-')
      a[i] = 62;
    else if (s[i] == '_')
      a[i] = 63;
  }
  long long sum = 1;
  for (int i = 1; i <= 63; i++) {
    int x = i;
    while (x) {
      b[i]++, x -= lowbit(x);
    }
  }
  for (int i = 0; i < n; i++) {
    sum = (sum * (1ll * bin[6 - b[a[i]]] % MOD)) % MOD;
  }
  printf("%I64d\n", sum % MOD);
  return 0;
}

```