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 DIV355C
{
	BufferedReader in;
	PrintWriter ob;
	StringTokenizer st;
	public static void main(String[] args) throws IOException {
		new DIV355C().run();
	}
	void run()throws IOException {
		in=new BufferedReader(new InputStreamReader(System.in));
		ob=new PrintWriter(System.out);
		solve();
		ob.flush();
	}
	void solve() throws IOException {
		int cnt[]=new int[64];
		for (int k=0; k<64 ; k++ ) {
			int res=0;			
			for (int i=0; i<64 ; i++ ) {
				for (int j=0; j<64 ; j++ ) {
					if((j&i)==k) {
						++res;
					}				
				}			
			}
			cnt[k]=res;
		}
		Hashtable<Character,Integer> ht=new Hashtable<Character,Integer>();
		char ch='0';
		for (int i=0; i<=9 ; i++ ) {
				ht.put(ch,i);
				ch+=1;
		}		
		 ch='A';
		for (int i=10; i<=35 ; i++ ) {
			ht.put(ch,i);
			ch+=1;
		}
		 ch='a';
		for (int i=36; i<=61 ; i++ ) {
			ht.put(ch,i);
			ch+=1;
		}
		ht.put('-',62);
		ht.put('_',63);
		char s[]=ns().toCharArray();
		long ans=1,mod=(int)1e9+7;
		for (int i=0; i<s.length ; i++ ) {
			int k=ht.get(s[i]);
			ans=((ans%mod)*1L*((cnt[k])%mod)%mod);
		}
		ob.println(ans);
	}
	String ns() throws IOException {
		return nextToken();
	}
	int ni()throws IOException {
		return Integer.parseInt(nextToken());
	}
	String nextToken() throws IOException {
		if(st==null || !st.hasMoreTokens())
			st=new StringTokenizer(in.readLine());
		return st.nextToken();
	}
}