codeforces 677c Vanya and Label



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 base64 and got new word. Now Vanya thinks of some strings and wants to know the number of pairs of words of length|s| (length of s), such that their bitwise AND is equal tos. 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 from0 to 9;
• letters from 'A' to 'Z' correspond to integers from10 to 35;
• letters from 'a' to 'z' correspond to integers from36 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 strings 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：就是说给你一个字符串，问你有多少个长度和它相同的字符串&后仍然和它等值，

思路：既然长度相同我们可以这样做去求每一个字符有多少个和它&一次值仍等于它，最后再把所有的值累乘不就好了么？关于怎么求每个字符有多少个和它&后值相等的呢？

首先如果是0的话，则有0&0,0&1,1&0三种，如果是1的话，就只有1&1这一种情况，好了，我们的字符在0-63之间，那就是说明每个字符二进制表示有6位呗，然后在对每一位来进行统计就ok了，

AC代码：

#include <iostream>#include <cstring>#include <cstdio>#include <cstdlib>#include <algorithm>#include <cstring>#include <cmath>using namespace std;char c[100005];const int mod = 1000000007;int main() {    cin >> c;    long long ans = 1;    int len = strlen(c);    for(int i=0; i<len; i++) {        int g;        if(c[i]>= '0' && c[i] <= '9') {            g = c[i]-'0';        }        if(c[i] >= 'A' && c[i] <= 'Z') {            g = c[i]-'A'+10;        }        if(c[i] >= 'a' && c[i] <= 'z') {            g = c[i]-'a'+36;        }        if(c[i] == '-') {            g  = 62;        }        if(c[i] == '_') {            g = 63;        }        for(int j = 0; j<6; j++) {            if(!((g >> j)&1)) {//右移操作得到的就是第j位是0还是1；                ans *= 3;                ans %= mod;            }        }    }    cout << ans << endl;    return 0;}