Codeforces Round #355 (Div. 2) C 预处理

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

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

题意：不同字符表示0~63   给你一个字符串判断 有多少对 长度相同的字符串的 &运算的结果 为所给的字符串

题解：模拟

       1&1=1   0&1=0  1&0=0  0&0=0

       打表预处理 0~63的满足情况的对数

       例如31=（11111）2

       共有三对    111111&011111

                      011111&111111

                      011111&011111

      结果乘积并取模

                 

#include<iostream>
#include<cstring>
#include<cstdio>
#include<queue>
#include<stack>
#include<cmath>
#include<map>
#define ll __int64 
#define pi acos(-1.0)
#define mod 1000000007
using namespace std;
map<char,int> mp;
ll a[100];
char b[100005];
ll quickmod(ll a,ll b)
{
    ll sum=1;
    while(b)
    {
        if(b&1)
            sum=(sum*a%mod);
        b>>=1;
        a=(a*a%mod);
    }
    return sum;
}
void init ()
{
    for(int i=0;i<=63;i++)
    {
        int n=i;
        int c=0;
         while (n >0)
      {
         if((n &1) ==1) 
            ++c ; 
         n >>=1 ; 
      }
      a[i]=quickmod(3,6-c);
    }
    int jishu=0;
    for(int i='0';i<='9';i++)
     {
      mp[i]=a[jishu];
      jishu++;
      }
    for(int i='A';i<='Z';i++)
    {
      mp[i]=a[jishu];
      jishu++;
    }
    for(int i='a';i<='z';i++)
      {
      mp[i]=a[jishu];
      jishu++;
      }
    mp['-']=a[62];
    mp['_']=a[63];
}
int main()
{
    init();
    ll ans=1;
    scanf("%s",b);
    int len=strlen(b);
    for(int i=0;i<len;i++)
        ans=(ans%mod*mp[b[i]])%mod;
    cout<<ans<<endl;
    return 0;
}