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.util.Scanner;

public class Label {
	public static void main(String[] args)
	{
		Scanner reader = new Scanner(System.in);
		String input = reader.nextLine();
		int n = input.length();
		int[] nums = new int[n];
		int numZeroes = 0;
		for (int i = 0; i < n;i++)
		{
			char temp = input.charAt(i);
			if (temp >= 48 && temp <= 57)
			{
				nums[i] = temp-48;
			}
			else if (temp >= 65 && temp <=90)
			{
				nums[i] = temp -55;
			}
			else if (temp >=97 && temp <= 122)
			{
				nums[i] = temp - 61;
			}
			else if (temp == 45)
			{
				nums[i] = 62;
			}
			else if (temp == 95)
			{
				nums[i] = 63;
			}
			numZeroes += (6-BitCount(nums[i]));
		}
		long accumulator = 1;
		for (int i = 0; i < numZeroes;i++)
		{
			accumulator *=3;
			if (accumulator > 1000000007)
			{
				accumulator = accumulator % 1000000007;
			}
		}
		System.out.println(accumulator);
	}
	public static int BitCount(int u)
	{
	     int uCount;

	     uCount = u - ((u >> 1) & 033333333333) - ((u >> 2) & 011111111111);
	     return ((uCount + (uCount >> 3)) & 030707070707) % 63;
	}
}
