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 double EPS = 1e-9;
const double PI = 3.14159265359;
vector<long long> primes;
void GeneratingPrimes(int top) {
  vector<bool> p(top + 1, 1);
  p[0] = p[1] = 0;
  for (int i = 2; i <= sqrt(top); i++) {
    if (!p[i]) continue;
    if (p[i]) {
      for (int j = i * i; j <= top; j += i) {
        p[j] = 0;
      }
    }
  }
  for (int i = 0; i < top + 1; i++)
    if (p[i]) primes.push_back(i);
}
long long gcd(long long a, long long b) { return (b == 0) ? a : gcd(b, a % b); }
long long POW(long long x, long long p) {
  long long ret = 1;
  while (p) {
    if (p & 1) ret *= x;
    x *= x;
    p /= 2;
  }
  return ret;
}
double dist(int a, int b, int x, int y) {
  return sqrt(pow(a - x, 2) + pow(b - y, 2));
}
int xadd[9] = {1, -1, 0, 1, -1, 0, 1, -1, 0};
int yadd[9] = {1, -1, 0, -1, 0, 1, 0, 1, -1};
map<char, int> m;
int MOD = 1e9 + 7;
void init() {
  for (int i = 0; i <= 9; i++) m[i + '0'] = i;
  for (int i = 10, j = 0; i <= 35; i++, j++) m['A' + j] = i;
  for (int i = 36, j = 0; i <= 61; i++, j++) m['a' + j] = i;
  m['-'] = 62;
  m['_'] = 63;
}
int main() {
  init();
  string s;
  cin >> s;
  int n = s.size();
  long long ans = 1;
  for (int i = 0; i < n; i++) {
    int x = m[s[i]];
    int c = 0;
    for (int a = 0; a < 64; a++) {
      for (int b = a; b < 64; b++)
        if ((a & b) == x && a != b) {
          c++;
        }
    }
    c *= 2;
    c++;
    ans *= c;
    ans %= MOD;
  }
  ans %= MOD;
  cout << ans;
  return 0;
}