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.
文字列$s$と$\&$を取った時に等しくなる$2$つの文字列の組み合わせいくつあるか？ $mod\ 10 ^9 + 7$で求める．
各桁についての組み合わせを出して掛けていった．組み合わせを全探索して組み合わせ数を出した． $\&と==$では$==$のほうが優先度が高いのでちゃんと括弧でくくる．
Code
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#include <iostream>
#include <vector>
#include <string>
#include <cstring>
#include <algorithm>
#include <sstream>
#include <map>
#include <set>
#include <bitset>
#define REP(i,k,n) for(int i=k;i<n;i++)
#define rep(i,n) for(int i=0;i<n;i++)
#define INF 1<<30
#define pb push_back
#define mp make_pair
#define MOD 1000000007
using namespace std ;
typedef long long ll ;
typedef pair < int , int > P ;
int main () {
vector < char > v ;
map < char , int > m ;
rep ( i , 10 ) {
m [ '0' + i ] = i ;
v . push_back ( '0' + i );
}
rep ( i , 26 ) {
m [ 'A' + i ] = 10 + i ;
m [ 'a' + i ] = 36 + i ;
v . push_back ( 'A' + i );
v . push_back ( 'a' + i );
}
m [ '-' ] = 62 ;
m [ '_' ] = 63 ;
v . push_back ( '-' );
v . push_back ( '_' );
map < char , ll > cnt ;
sort ( v . begin (), v . end ());
rep ( i , v . size ()) {
rep ( j , v . size ()) {
rep ( k , v . size ()) {
if (( m [ v [ j ]] & m [ v [ k ]]) == m [ v [ i ]]) {
cnt [ v [ i ]] ++ ;
}
}
}
}
ll ans = 1 ;
string s ;
cin >> s ;
rep ( i , s . size ()) {
ans *= cnt [ s [ i ]];
ans %= MOD ;
}
cout << ans << endl ;
return 0 ;
}