Codeforces 677C Vanya and Label【思维】

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本文介绍了一道关于字符串匹配的问题，需要统计与给定字符串等长且按位与运算后值相同的字符串数量。通过分析字符对应的数值及二进制表示，采用高效算法求解并给出AC代码。

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

题目大意：

给你一个字符串，并且告诉你每个字符的价值，让你统计一共有多少个字符串，和原字符串等长，并且每一位的&之后的值还是原字符串的值。

思路：

对于一个二进制位来讲，如果一个位子是1.那么这个字符进行替换的字符这个位子肯定也是1才行。

那么如果一个位子是0.那么这个字符进行替换的字符这个位子可能是三种情况：0&1 1&0 0&0；

那么我们的任务就是统计一个位子上的字符的二进制有多少个0.有一个乘一个3即可。

Ac代码：

#include<stdio.h>
#include<string.h>
#include<math.h>
using namespace std;
#define mod 1000000007
#define ll __int64
char a[100700];
ll vis[100];
int judge(char a)
{
    if(a>='0'&&a<='9')return a-'0';
    if(a>='A'&&a<='Z')return a-'A'+10;
    if(a>='a'&&a<='z')return a-'a'+36;
    if(a=='-')return 62;
    if(a=='_')return 63;
}
int main()
{
    while(~scanf("%s",a))
    {
        ll output=1;
        int n=strlen(a);
        for(int i=0;i<n;i++)
        {
            int tmp=judge(a[i]);
            int cnt=0;
            while(tmp)
            {
                cnt++;
                if(tmp%2==0)
                output*=3;
                tmp/=2;
                output%=mod;
            }
            if(cnt<6)
            {
                while(cnt<6)output*=3,cnt++,output%=mod;
            }
        }
        printf("%I64d\n",output);
    }
}