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;
char READ_DATA;
int SIGNAL_INPUT;
template <typename Type>
inline char ru(Type &v) {
  SIGNAL_INPUT = 1;
  while ((READ_DATA = getchar()) < '0' || READ_DATA > '9')
    if (READ_DATA == '-')
      SIGNAL_INPUT = -1;
    else if (READ_DATA == EOF)
      return EOF;
  v = READ_DATA - '0';
  while ((READ_DATA = getchar()) >= '0' && READ_DATA <= '9')
    v = v * 10 + READ_DATA - '0';
  v *= SIGNAL_INPUT;
  return 1;
}
template <typename A, typename B>
inline char ru(A &x, B &y) {
  if (ru(x) == EOF) return EOF;
  ru(y);
  return 2;
}
template <typename A, typename B, typename C>
inline char ru(A &x, B &y, C &z) {
  if (ru(x) == EOF) return EOF;
  ru(y);
  ru(z);
  return 3;
}
template <typename A, typename B, typename C, typename D>
inline char ru(A &x, B &y, C &z, D &w) {
  if (ru(x) == EOF) return EOF;
  ru(y);
  ru(z);
  ru(w);
  return 4;
}
struct Edge {
  int u, v, next;
  int w, cap, flow;
  Edge(int _u = 0, int _v = 0, int nxt = -1, int _w = 1, int _cap = 0) {
    u = _u;
    v = _v;
    w = _w;
    cap = _cap;
    flow = 0;
    next = nxt;
  }
};
const int mod = 1e9 + 7;
const int maxn = 1e5 + 2;
int val(char c) {
  if (c >= '0' && c <= '9') return c - '0';
  if (c >= 'A' && c <= 'Z') return 10 + c - 'A';
  if (c >= 'a' && c <= 'z') return 36 + c - 'a';
  if (c == '-') return 62;
  if (c == '_') return 63;
}
int bitcnt(int x) {
  int ans = 0;
  while (x) {
    if (x & 1) ++ans;
    x >>= 1;
  }
  return 6 - ans;
}
long long ans = 1;
char s[maxn];
long long qkpow(long long a, long long b) {
  long long x = a;
  a = 1;
  while (b) {
    if (b & 1) (a *= x) %= mod;
    (x *= x) %= mod;
    b >>= 1;
  }
  return a;
}
long long tot;
int main() {
  scanf("%s", s);
  for (int i = 0; s[i]; ++i) {
    tot += bitcnt(val(s[i]));
  }
  ans = qkpow(3, tot);
  printf("%I64d", ans);
}
