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
#include <bits/stdc++.h>
using namespace std;
const long long int mod = 1e9 + 7;
long long int fact[300010];
long long int power(long long int x, long long int n) {
  long long int res = 1;
  while (n > 0) {
    if (n % 2 == 1) res = (res * x) % mod;
    x = (x * x) % mod;
    n /= 2;
  }
  return res % mod;
}
void factorial() {
  fact[0] = fact[1] = 1;
  for (long long int i = 2; i < 300005; i++) fact[i] = (fact[i - 1] * i) % mod;
}
long long int inv(long long int n) { return power(n, mod - 2) % mod; }
long long int comb(long long int n, long long int k) {
  long long int ans = 1;
  ans = (ans * fact[n]) % mod;
  ans = (ans * power(fact[n - k], mod - 2)) % mod;
  ans = (ans * power(fact[k], mod - 2)) % mod;
  return ans;
}
long long int solve(long long int x, long long int n) {
  long long int ans = 0;
  for (long long int i = 2; (i * i * i) <= x; i++) {
    ans += (x / (i * i * i));
  }
  return (ans);
}
long long int check(long long int x) {
  long long int ans = 0;
  while (x > 0) {
    ans += (x % 2);
    x /= 2;
  }
  return (6 - ans);
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  factorial();
  long long int t;
  t = 1;
  while (t--) {
    string s;
    cin >> s;
    map<char, long long int> mp;
    for (long long int i = 0; i <= 9; i++) {
      mp[i + '0'] = check(i);
    }
    long long int i, j, k;
    char c = 'A';
    for (i = 10; i <= 35; i++) {
      mp[c] = check(i);
      c++;
    }
    c = 'a';
    for (i = 36; i <= 61; i++) {
      mp[c] = check(i);
      c++;
    }
    mp['-'] = check(62);
    mp['_'] = check(63);
    long long int ans = 1;
    for (i = 0; i < s.size(); i++) {
      ans = (ans * (power(3, mp[s[i]]))) % mod;
    }
    cout << ans;
  }
}
