#include <bits/stdc++.h>
using namespace std;
static const int INF = 0x3f3f3f3f;
static const long long INFL = 0x3f3f3f3f3f3f3f3fLL;
template <typename T, typename U>
static void amin(T &x, U y) {
  if (y < x) x = y;
}
template <typename T, typename U>
static void amax(T &x, U y) {
  if (x < y) x = y;
}
template <int MOD>
struct ModInt {
  static const int Mod = MOD;
  unsigned x;
  ModInt() : x(0) {}
  ModInt(signed sig) {
    int sigt = sig % MOD;
    if (sigt < 0) sigt += MOD;
    x = sigt;
  }
  ModInt(signed long long sig) {
    int sigt = sig % MOD;
    if (sigt < 0) sigt += MOD;
    x = sigt;
  }
  int get() const { return (int)x; }
  ModInt &operator+=(ModInt that) {
    if ((x += that.x) >= MOD) x -= MOD;
    return *this;
  }
  ModInt &operator-=(ModInt that) {
    if ((x += MOD - that.x) >= MOD) x -= MOD;
    return *this;
  }
  ModInt &operator*=(ModInt that) {
    x = (unsigned long long)x * that.x % MOD;
    return *this;
  }
  ModInt operator+(ModInt that) const { return ModInt(*this) += that; }
  ModInt operator-(ModInt that) const { return ModInt(*this) -= that; }
  ModInt operator*(ModInt that) const { return ModInt(*this) *= that; }
};
int main() {
  string s;
  while (cin >> s) {
    ModInt<1000000007> ans = 1;
    for (char c : s) {
      int x;
      if (c == '-')
        x = 62;
      else if (c == '_')
        x = 63;
      else if (c <= '9')
        x = c - '0';
      else if (c <= 'Z')
        x = c - 'A' + 10;
      else
        x = c - 'a' + 36;
      int p = 0;
      for (int(i) = 0; (i) < (int)(6); ++(i))
        if (~x >> i & 1) ans *= 3;
    }
    printf("%d\n", ans.get());
  }
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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;
static const int INF = 0x3f3f3f3f;
static const long long INFL = 0x3f3f3f3f3f3f3f3fLL;
template <typename T, typename U>
static void amin(T &x, U y) {
  if (y < x) x = y;
}
template <typename T, typename U>
static void amax(T &x, U y) {
  if (x < y) x = y;
}
template <int MOD>
struct ModInt {
  static const int Mod = MOD;
  unsigned x;
  ModInt() : x(0) {}
  ModInt(signed sig) {
    int sigt = sig % MOD;
    if (sigt < 0) sigt += MOD;
    x = sigt;
  }
  ModInt(signed long long sig) {
    int sigt = sig % MOD;
    if (sigt < 0) sigt += MOD;
    x = sigt;
  }
  int get() const { return (int)x; }
  ModInt &operator+=(ModInt that) {
    if ((x += that.x) >= MOD) x -= MOD;
    return *this;
  }
  ModInt &operator-=(ModInt that) {
    if ((x += MOD - that.x) >= MOD) x -= MOD;
    return *this;
  }
  ModInt &operator*=(ModInt that) {
    x = (unsigned long long)x * that.x % MOD;
    return *this;
  }
  ModInt operator+(ModInt that) const { return ModInt(*this) += that; }
  ModInt operator-(ModInt that) const { return ModInt(*this) -= that; }
  ModInt operator*(ModInt that) const { return ModInt(*this) *= that; }
};
int main() {
  string s;
  while (cin >> s) {
    ModInt<1000000007> ans = 1;
    for (char c : s) {
      int x;
      if (c == '-')
        x = 62;
      else if (c == '_')
        x = 63;
      else if (c <= '9')
        x = c - '0';
      else if (c <= 'Z')
        x = c - 'A' + 10;
      else
        x = c - 'a' + 36;
      int p = 0;
      for (int(i) = 0; (i) < (int)(6); ++(i))
        if (~x >> i & 1) ans *= 3;
    }
    printf("%d\n", ans.get());
  }
  return 0;
}

```