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 JAVA):

Solution:
import java.io.*;
import java.util.*;
import java.math.*;

public class C677 {

    public static BufferedReader read = null;
    public static PrintWriter out = null;
    public static StringTokenizer token = null;


    public static void solve()
    {

        String s = next();
        int[] ways = new int[64];
        for(int i=0; i<64; i++)
        {
            for(int j = 0; j<64; j++)
            {
                //ways[i] = # of ways to make i from j,k such that j & k = i
                ways[i & j]++;
            }
        }
        long mod = 1000000000+7;
        long tot = 1;
        for(int i=0; i<s.length(); i++)
        {
            int tar = 0;
            char c = s.charAt(i);
            if(Character.isDigit(c))
            {
                tar = (int)(c - '0');
            }
            else if(Character.isAlphabetic(c))
            {
                if(Character.isUpperCase(c))
                {
                    tar = (int)(c - 'A' + 10);
                }
                else
                {
                    tar = (int)(c - 'a' + 36);
                }
            }
            else
            {
                if(c == '-') tar = 62;
                else if(c == '_') tar = 63;
            }

            tot *= ways[tar];
            tot %= mod;
        }
        out.println(tot % mod);

    }

    public static void main(String[] args)
    {
        read = new BufferedReader(new InputStreamReader(System.in));
        out = new PrintWriter(System.out);
        solve();
        out.flush();
        out.close();

    }

    // i/o functions
    public static String next() // returns the next string
    {
        while(token == null || !token.hasMoreTokens())
        {
            try {
                token = new StringTokenizer(read.readLine());
            } catch (IOException e) {
                // TODO Auto-generated catch block
                e.printStackTrace();
            }
        }
        return token.nextToken();
    }

    public static int nint()
    {
        return Integer.parseInt(next());
    }
    public static long nlong()
    {
        return Long.parseLong(next());
    }
    public static double ndouble()
    {
        return Double.parseDouble(next());
    }
    public static int[] narr(int n)
    {
        int[] a = new int[n];
        for(int i=0; i<n; i++) a[i] = nint();
        return a;
    }

    public static long[] nal(int n)
    {
        long[] a = new long[n];
        for(int i=0; i<n; i++) a[i] = nlong();
        return a;
    }


}