codeforces 355 div2 C. Vanya and Label 水题

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 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
思路：打表得到并之后的种数；

#include<bits/stdc++.h>
using namespace std;
#define ll __int64
#define mod 1000000007
#define inf 999999999
#define pi (4*atan(1.0))
const int N=1e2+10,M=1e5+10;
ll flag[N];
ll getnumber(char a)
{
    if(a=='-')
    return 62;
    if(a=='_')
    return 63;
    if(a>='0'&&a<='9')
    return a-'0';
    if(a>='a'&&a<='z')
    return a-'a'+36;
    if(a>='A'&&a<='Z')
    return a-'A'+10;
}
char a[100010];
int main()
{
    ll x,y,z,i,t;
    for(i=0;i<=63;i++)
    for(t=0;t<=63;t++)
    {
        ll gg=(i&t);
        if(gg<64)
        flag[gg]++;
    }
    scanf("%s",a);
    x=strlen(a);
    ll ans=1;
    for(i=0;i<x;i++)
    {
        ans*=flag[getnumber(a[i])];
        ans%=mod;
    }
    printf("%I64d\n",ans);
    return 0;
}