#include <bits/stdc++.h>
using namespace std;
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long lcm(long long a, long long b) { return (a / gcd(a, b)) * b; }
int value(char c) {
  if (c <= '9' && c >= '0') return c - '0';
  if (c >= 'A' && c <= 'Z') return c - 'A' + 10;
  if (c >= 'a' && c <= 'z') return c - 'a' + 36;
  if (c == '_') return 63;
  return 62;
}
int set_bits(int n) {
  int count = 0;
  while (n) {
    count += n & 1;
    n >>= 1;
  }
  return count;
}
long long pow(int a, int b, int MOD) {
  long long x = 1, y = a;
  while (b > 0) {
    if (b % 2 == 1) {
      x = (x * y);
      if (x > MOD) x %= MOD;
    }
    y = (y * y);
    if (y > MOD) y %= MOD;
    b /= 2;
  }
  return x;
}
long long InverseEuler(int n, int MOD) { return pow(n, MOD - 2, MOD); }
vector<long long> f(600001, 1);
long long C(int n, int r, int MOD) {
  return (f[n] *
          ((InverseEuler(f[r], MOD) * InverseEuler(f[n - r], MOD)) % MOD)) %
         MOD;
}
int main() {
  long long i, j, h, k, m, t, ans, n, p, flag, q, sum;
  string s;
  cin >> s;
  ans = 0;
  for (i = 2; i <= 600000; i++) f[i] = (f[i - 1] * i) % 1000000007;
  for (i = 0; i < s.length(); i++) {
    k = value(s[i]);
    j = 6 - set_bits(k);
    ans += j;
  }
  j = 2;
  t = ans;
  k = 1;
  for (i = 1; i <= ans; i++) {
    k += j * t;
    t = C(ans, i + 1, 1000000007);
    j *= 2;
    j = j % 1000000007;
    k = k % 1000000007;
  }
  cout << k << endl;
  return 0;
}

### Prompt

Please create a solution in cpp to the following problem:

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 gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long lcm(long long a, long long b) { return (a / gcd(a, b)) * b; }
int value(char c) {
  if (c <= '9' && c >= '0') return c - '0';
  if (c >= 'A' && c <= 'Z') return c - 'A' + 10;
  if (c >= 'a' && c <= 'z') return c - 'a' + 36;
  if (c == '_') return 63;
  return 62;
}
int set_bits(int n) {
  int count = 0;
  while (n) {
    count += n & 1;
    n >>= 1;
  }
  return count;
}
long long pow(int a, int b, int MOD) {
  long long x = 1, y = a;
  while (b > 0) {
    if (b % 2 == 1) {
      x = (x * y);
      if (x > MOD) x %= MOD;
    }
    y = (y * y);
    if (y > MOD) y %= MOD;
    b /= 2;
  }
  return x;
}
long long InverseEuler(int n, int MOD) { return pow(n, MOD - 2, MOD); }
vector<long long> f(600001, 1);
long long C(int n, int r, int MOD) {
  return (f[n] *
          ((InverseEuler(f[r], MOD) * InverseEuler(f[n - r], MOD)) % MOD)) %
         MOD;
}
int main() {
  long long i, j, h, k, m, t, ans, n, p, flag, q, sum;
  string s;
  cin >> s;
  ans = 0;
  for (i = 2; i <= 600000; i++) f[i] = (f[i - 1] * i) % 1000000007;
  for (i = 0; i < s.length(); i++) {
    k = value(s[i]);
    j = 6 - set_bits(k);
    ans += j;
  }
  j = 2;
  t = ans;
  k = 1;
  for (i = 1; i <= ans; i++) {
    k += j * t;
    t = C(ans, i + 1, 1000000007);
    j *= 2;
    j = j % 1000000007;
    k = k % 1000000007;
  }
  cout << k << endl;
  return 0;
}

```