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
import java.io.*;
import java.util.*;

public class Main {

    public final static int C = 6, MOD = (int)1e9 + 7;

    int p3[] = new int[10];

    int getNum(int mask) {
        return p3[C - Integer.bitCount(mask)];
    }
    int convert(char c) {
        if (Character.isDigit(c)) {
            return c - '0';
        } else if (Character.isUpperCase(c)) {
            return 10 + c - 'A';
        } else if (Character.isLowerCase(c)) {
            return 36 + c - 'a';
        }
        return c == '-' ? 62 : 63;
    }
    public void solve() throws IOException {
        p3[0] = 1;
        for (int i = 1 ; i < 10 ; i++) {
            p3[i] = (p3[i-1] * 3) % MOD;
        }

        String s = next();
        long res = 1;
        for (int i = 0 ; i < s.length() ; i++) {
            char c = s.charAt(i);
            int m = convert(c);
            res = (1L * res * getNum(m)) % MOD;
        }
        out.println(res);
    }
    

    
    
    BufferedReader bf;
    StringTokenizer st;
    PrintWriter out;
    
    public String next() throws IOException {
        while (st == null || !st.hasMoreTokens()) 
            st = new StringTokenizer(bf.readLine());
        return st.nextToken();
    }
    public int nextInt() throws IOException {
        return Integer.parseInt(next());
    }
    
    public long nextLong() throws IOException {
        return Long.parseLong(next());
    }

    public Main() throws IOException {
        bf = new BufferedReader(new InputStreamReader(System.in));
        out = new PrintWriter(new OutputStreamWriter(System.out));		
        solve();
        bf.close();
        out.close();
    }
    
    public static void main(String args[]) throws IOException {
        new Main();
    }
}