Codeforces Round #355 (Div. 2) C 预处理

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

output

input

V_V

output

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:

z&_ = 61&63 = 61 = z
_&z = 63&61 = 61 = z
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

 1 #include<iostream>
 2 #include<cstring>
 3 #include<cstdio>
 4 #include<queue>
 5 #include<stack>
 6 #include<cmath>
 7 #include<map>
 8 #define ll __int64 
 9 #define pi acos(-1.0)
10 #define mod 1000000007
11 using namespace std;
12 map<char,int> mp;
13 ll a[100];
14 char b[100005];
15 ll quickmod(ll a,ll b)
16 {
17     ll sum=1;
18     while(b)
19     {
20         if(b&1)
21             sum=(sum*a%mod);
22         b>>=1;
23         a=(a*a%mod);
24     }
25     return sum;
26 }
27 void init ()
28 {
29     for(int i=0;i<=63;i++)
30     {
31         int n=i;
32         int c=0;
33          while (n >0)
34       {
35          if((n &1) ==1) 
36             ++c ; 
37          n >>=1 ; 
38       }
39       a[i]=quickmod(3,6-c);
40     }
41     int jishu=0;
42     for(int i='0';i<='9';i++)
43      {
44       mp[i]=a[jishu];
45       jishu++;
46       }
47     for(int i='A';i<='Z';i++)
48     {
49       mp[i]=a[jishu];
50       jishu++;
51     }
52     for(int i='a';i<='z';i++)
53       {
54       mp[i]=a[jishu];
55       jishu++;
56       }
57     mp['-']=a[62];
58     mp['_']=a[63];
59 }
60 int main()
61 {
62     init();
63     ll ans=1;
64     scanf("%s",b);
65     int len=strlen(b);
66     for(int i=0;i<len;i++)
67         ans=(ans%mod*mp[b[i]])%mod;
68     cout<<ans<<endl;
69     return 0;
70 }