Codeforces Round #355 (Div. 2)C. Vanya and Label

C. Vanya and Label

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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 modulo109 + 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

/* ***********************************************Author       : rycCreated Time : 2016-08-03 WednesdayFile Name    : E:\acmcode\codeforces\355div2C.cppLANGUAGE     : c++Copyright 2016 ryc All Rights Reserved************************************************ */#include<iostream>#include<cstdio>#include<cstdlib>#include<cstring>#include<algorithm>#include<list>#include<vector>#include<queue>#include<cmath>#include<map>#define MOD 1000000007using namespace std;const int maxn=100010;char str[maxn];long long num[maxn];int main(){    for(int i=0;i<64;++i){        for(int j=0;j<64;++j){            num[i&j]++;//打表求出i&j这个字符有可有多少对i j构成        }    }    scanf("%s",str);    long long ans=1;    for(int i=0;str[i];++i){        int temp;        if(str[i]>='0'&&str[i]<='9'){            temp=str[i]-'0';        }        else if(str[i]>='A'&&str[i]<='Z'){            temp=str[i]-'A'+10;        }        else if(str[i]>='a'&&str[i]<='z'){            temp=str[i]-'a'+36;        }        else if(str[i]=='-'){            temp=62;        }        else if(str[i]=='_'){            temp=63;        }        ans=ans*(num[temp])%MOD;    }    printf("%lld\n",ans%MOD);    return 0;}