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.util.*;

public class Solution {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        String s = sc.next();
        int[] count = new int[64];
        for (char c : s.toCharArray()) {
            if (c >= '0' && c <= '9')
                count[c - '0']++;
            else if (c >= 'A' && c <= 'Z')
                count[c - 'A' + 10]++;
            else if (c >= 'a' && c <= 'z')
                count[c - 'a' + 36]++;
            else if (c == '-')
                count[62]++;
            else
                count[63]++;
        }

        int countZero = 0;
        for (int i = 0; i < 64; i++) {
            int countZeroI = 0;
            for (int j = 0; j < 6; j++) {
                if ((i & (1 << j)) == 0)
                    countZeroI++;
            }
            countZero += count[i] * countZeroI;
        }

        long res = 1;
        for (int i = 0; i < countZero; i++) {
            res = (res * 3) % 1000000007;
        }
        // System.out.println(countZero);
        System.out.println(res);
    }

}