Beginner and find -4:6 a friend

Title Description Description

There is a saying: Know 6 people, you know people all over the world.

Aiden now has a diagram of the situation in which n people know each other. Aiden wants to know if he can only know 6 people and indirectly know these n individuals?

Enter a description Input Description

The first line, two numbers n,m, indicates that there are n individuals, M pairs of cognitive relationships.

The next M-line, which is two ai,bi per line, indicates that AI and bi are aware of each other.

There is no guarantee that the cognition relationship does not recur and ai≠bi.

n the number of individuals is 1 ... N.

Output description Output Description

If you know only 6 people can indirectly know this n person, then output "^_^".

If not, then the first line output "t_t", the second row of the output of the 6 people can know the most indirect number of people.

The output does not include quotation marks.

Sample input Sample Input

6 7

1 2

1 3

2 4

3 5

4 6

5 6

3 2

Sample output Sample Output

^_^

Data range and Tips Data Size & Hint

For 30% of data, ensure 0

For 50% of data, ensure 0

For 100% of data, ensure 0

Type R=record num,v:longint;    End;var I,j,k,l:longint;    Father:array[1..10000]of Longint;    N,m:longint;    X,y:longint;    Cost:array[1..10000]of R;    Num:array[1..10000]of integer;         Used:array[1..10000]of boolean;function Find (x:longint): Longint;               Begin if Father[x]=x then exit (x);               Father[x]:=find (Father[x]);         Exit (Father[x]);          End;procedure Union (X,y:longint);          Begin Father[find (x)]:=find (Father[y]);          End;procedure Choose;              var i,j,k:longint;          Max,maxx:longint;                          Begin for I:=1 to 6 does begin max:=0;  For J:=1 to N does if (Cost[j].num>max) and (not USED[J]) then                                            Begin Max:=cost[j].num;                                      Maxx:=j;                          End                          Used[maxx]:=true;            L:=l+max;        End      End;begin readln (n,m);      Fillchar (Father,sizeof (father), 0);      For I:=1 to n do father[i]:=i;                For I:=1 to M do begin READLN (x, y);          If find (x) <>find (Y) then union (x, y);      End      Fillchar (cost,sizeof (cost), 0);      Fillchar (num,sizeof (num), 0);      Fillchar (used,sizeof (used), false);      For I:=1 to N Do Inc (Num[find (i)]);      k:=0;                        For I:=1 to N does if Num[i]<>0 then BEGIN Inc (K);                        Cost[k].v:=i;                  Cost[k].num:=num[i];      End      l:=0;                    If K<=6 then begin Writeln (' ^_^ ');              Halt                    end else begin choose;                    Writeln (' t_t ');              Writeln (l); End;end.

Beginner and check -4:6 a friend