A - deep sea exploration

A long, long time ago , A beautiful man came to the seaside , The sea was stormy . The beautiful man hopes to find a mermaid in the sea , But unfortunately he only found the octopus .

 

However , On the other side of the world , People are actively collecting information about monster behavior , In order to develop powerful weapons to deal with the octopus monster . Due to the frequent occurrence of earthquakes , And bad weather , So our satellites can't locate monsters well , So they can't hit the target well . The analysis results of the first shot will be reflected in a picture by n Points and m On an undirected graph composed of edges . Now let's determine whether this picture can be regarded as an octopus monster .

 

For the sake of simplicity , We assume that the shape of the octopus monster is like this , He has a spherical body , Then there are many tentacles attached to his body . It can be represented as an undirected graph , In the graph, it can be considered as three or more trees （ Representative tentacle ） form , The roots of these trees are in a ring in the graph （ This ring represents a spherical body ）.

 

Title assurance , There are no duplicate edges in the graph , There is no self link .

 

Input

A single set of test data  
The first line gives two numbers ,n Represents the number of points in the graph ,m Represents the number of edges in the graph . （1≤ n≤100,0≤ m≤ n*(n-1)/2 ） 
Next m Rows give information about edges , 
Each line has two upper numbers x,y (1≤ x,y≤ n,x≠y) 
Indication point x Sum point y There are edges between them . At most one edge of each pair of points is connected , Point itself will not have edge to itself .

Output

All in one line , If the given graph is considered to be an octopus, output "FHTAGN!"( No quotes ), Otherwise output "NO"( No quotes ).

Sample Input

6 6
6 3
6 4
5 1
2 5
1 4
5 4

Sample Output

FHTAGN!

The question ： Give two numbers n,m, On behalf of n Nodes m side , And then follow m Two numbers per row a,b representative a Follow b Connected to a . Ask if the picture is an octopus , The shape of an octopus is that it has a ring and can form a connected graph . Yes, output FHTAGN!, Otherwise output NO.

Ideas ： This problem only needs to satisfy two conditions to form an octopus ,1. Construct connected graph .2. There is only one ring . Using parallel search set to realize , Merge during input , If two nodes have a common ancestor, it means that a ring is formed , Then add one to the number of rings . And judge how many ancestors there are after the merger , If there is a ring and only one ancestor , It can form the shape of an octopus . The code is as follows ：

#include<stdio.h>
#include<string.h>
int f[110];
int getf(int v)
{
    if(f[v]==v)
        return v;
    else
    {
        f[v]=getf(f[v]);
        return f[v];
    }
}
int merge(int u,int v)
{
    int t1,t2;
    t1=getf(u);
    t2=getf(v);
    if(t1!=t2)
    {
        f[t2]=t1;
        return 0;
    }
    return 1;
}
int main()
{
    int i,x,y,n,m,r,sum;
    while(~scanf("%d %d",&n,&m))
    {
        r=0;
        sum=0;
        for(i=1; i<=n; i++)
            f[i]=i;
        for(i=0; i<m; i++)
        {
            scanf("%d %d",&x,&y);
            if(merge(x,y)==1)
                r++;
        }
        for(i=1; i<=n; i++)
        {
            if(f[i]==i)
                sum++;
        }
        if(sum==1&&r==1)
            printf("FHTAGN!\n");
        else
            printf("NO\n");
    }
    return 0;
}