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 bool testcases = false;
int64_t seed;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
inline int64_t rnd(int64_t l = 0, int64_t r = INT_MAX) {
  return uniform_int_distribution<int64_t>(l, r)(rng);
}
inline bool in_range(int64_t a, int64_t l, int64_t r) {
  return (a >= l && a <= r);
}
template <typename H, typename... T>
void inp(H &head) {
  cin >> head;
}
template <typename H, typename... T>
void inp(H &head, T &...tail) {
  cin >> head;
  inp(tail...);
}
template <typename T>
bool isPrime(T n) {
  for (T i = 2; i * i <= n; i++)
    if (n % i == 0) return false;
  return n >= 2;
}
template <typename T>
inline istream &operator>>(istream &in, vector<T> &a) {
  for (T &x : a) in >> x;
  return in;
}
template <typename T, typename U>
inline istream &operator>>(istream &in, pair<T, U> &a) {
  in >> a.first >> a.second;
  return in;
}
void solve();
void fast_io();
int32_t main(int32_t argc, char **argv) {
  fast_io();
  int64_t TC = 1;
  if (testcases) cin >> TC;
  for (int64_t tc = 1; tc <= TC; tc++) solve(), 108;
  return 0;
}
map<char, int64_t> mp;
const int64_t mod = 1e9 + 7;
int64_t mul(int64_t a, int64_t b) { return ((a % mod) * (b % mod)) % mod; }
void solve() {
  string s;
  cin >> s;
  int64_t n = s.size();
  for (int64_t i = '0'; i <= '9'; i++) mp[i] = i - '0';
  for (int64_t i = 'A'; i <= 'Z'; i++) mp[i] = i - 'A' + 10;
  for (int64_t i = 'a'; i <= 'z'; i++) mp[i] = i - 'a' + 36;
  mp['-'] = 62;
  mp['_'] = 63;
  vector<int64_t> a(n);
  for (int64_t i = 0; i < n; i++) a[i] = mp[s[i]];
  int64_t ans = 1;
  for (int64_t x : a) {
    int64_t cnt = 0;
    for (int64_t i = 0; i < 64; i++) {
      for (int64_t j = 0; j < 64; j++) {
        if ((i & j) == x) cnt++;
      }
    }
    ans = mul(ans, cnt);
  }
  cout << ans << '\n';
}
void fast_io() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
}