Vanya and Label

Description

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.

Sample Input

Input

z

Output

3

Input

V_V

Output

9

Input

Codeforces

Output

130653412

#include<map>#include<set>#include<cmath>#include<queue>#include<bitset>#include<math.h>#include<vector>#include<string>#include<stdio.h>#include<cstring>#include<iostream>#include<algorithm>using namespace std;const int maxn=100000+10;const int modul=1000000000+7;char st[maxn];int pan(int n){int i;int num=0;for(i=0;i<=5;i++){int x=n>>i;if(!(x&1))num++;}return num;}int tra(char a){int n;if(a>='0'&&a<='9'){n=a-'0';}else if(a>='A'&&a<='Z'){n=a-'A'+10;}else if(a>='a'&&a<='z'){n=a-'a'+36;}else if(a=='-'){n=62;}else if(a=='_'){n=63;}return pan(n);}int main(){int ans=0;scanf("%s",&st);int n=strlen(st);int i;__int64 j=1;for(i=0;i<n;i++){ans+=tra(st[i]);}for(i=0;i<ans;i++){j=(j*3)%(modul);}printf("%I64d\n",j);return 0;}