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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const long double PI = 3.1415926535897932384626;
const long long int MAX = 1000000000 + 7;
const int INF = 0x3f3f3f3f;
inline long long int add(long long int a, long long int b, long long int mod) {
  long long int c = (long long int)((long long int)(a) + (long long int)(b));
  while (c >= mod) {
    c -= mod;
  };
  while (c < 0) {
    c += mod;
  };
  return c;
}
inline long long int sub(long long int a, long long int b, long long int mod) {
  return add(a, mod - b, mod);
}
inline long long int mul(long long int a, long long int b, long long int mod) {
  long long int c = (long long int)((long long int)(a) * (long long int)(b));
  c %= mod;
  while (c < 0) {
    c += mod;
  };
  return c;
}
inline long long int mod_pow(long long int a, long long int n,
                             long long int b) {
  long long int res = 1;
  while (n) {
    if (n & 1) {
      res = mul(res, a, b);
    }
    a = mul(a, a, b);
    n >>= 1;
  }
  return add(res, 0, b);
}
inline long long int mod_div(long long int a, long long int b,
                             long long int mod) {
  long long int ans = mul(a, mod_pow(b, mod - 2, mod), mod);
  return ans;
}
long long int cnt[100 + 1];
void calc() {
  for (int i = 0; i <= 63; i++) {
    for (int j = 0; j <= 63; j++) {
      cnt[i & j]++;
    }
  }
}
int main(int argc, char const *argv[]) {
  calc();
  string a;
  cin >> a;
  long long int ans = 1;
  for (int i = 0; i < a.size(); i++) {
    if (a[i] == '-') {
      int by = 62;
      ans = mul(ans, cnt[by], MAX);
    } else if (a[i] == '_') {
      int by = 63;
      ans = mul(ans, cnt[by], MAX);
    } else if (a[i] >= '0' and a[i] <= '9') {
      int by = a[i] - '0';
      ans = mul(ans, cnt[by], MAX);
    } else if (a[i] >= 'a' and a[i] <= 'z') {
      int by = a[i] - 'a' + 36;
      ans = mul(ans, cnt[by], MAX);
    } else if (a[i] >= 'A' and a[i] <= 'Z') {
      int by = a[i] - 'A' + 10;
      ans = mul(ans, cnt[by], MAX);
    }
  }
  cout << ans << endl;
  return 0;
}