Problem:
Score : 300 points

Problem Statement

In Takahashi Kingdom, there is an archipelago of N islands, called Takahashi Islands.
For convenience, we will call them Island 1, Island 2, ..., Island N.

There are M kinds of regular boat services between these islands.
Each service connects two islands. The i-th service connects Island a_i and Island b_i.

Cat Snuke is on Island 1 now, and wants to go to Island N.
However, it turned out that there is no boat service from Island 1 to Island N, so he wants to know whether it is possible to go to Island N by using two boat services.

Help him.

Constraints

3 ≤ N ≤ 200 000

1 ≤ M ≤ 200 000

1 ≤ a_i < b_i ≤ N

(a_i, b_i) \neq (1, N)

If i \neq j, (a_i, b_i) \neq (a_j, b_j).

Input

Input is given from Standard Input in the following format:

N M
a_1 b_1
a_2 b_2
:
a_M b_M

Output

If it is possible to go to Island N by using two boat services, print POSSIBLE; otherwise, print IMPOSSIBLE.

Sample Input 1

3 2
1 2
2 3

Sample Output 1

POSSIBLE

Sample Input 2

4 3
1 2
2 3
3 4

Sample Output 2

IMPOSSIBLE

You have to use three boat services to get to Island 4.

Sample Input 3

100000 1
1 99999

Sample Output 3

IMPOSSIBLE

Sample Input 4

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

Sample Output 4

POSSIBLE

You can get to Island 5 by using two boat services: Island 1 -> Island 4 -> Island 5.

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<iostream>
#include<cstring>
#include<cmath>
#include<cstdio>
using namespace std;

struct Edge
{
    int to;
    int next;
}a[400009];
int cnt;
int head[400009];
int n;
int vis[400009];
void addedg(int u,int v)
{
    a[cnt].to=v;
    a[cnt].next=head[u];
    head[u]=cnt++;
}
int flag=0;
void dfs(int s,int t)
{
    if(t>2)return;
    else if(t==2)
    {   if(s==n)
         {
             flag=1;
         }
        return;
    }
    else
    {
        for(int e=head[s];e!=-1;e=a[e].next)
        {
            /*if(vis[e]==0)
            {
                vis[e]=1;
                dfs(a[e].to,t+1);
                vis[e]=0;
            }*/
            dfs(a[e].to,t+1);
        }
    }
}
int main()
{
   int m;
   cin>>n>>m;
   memset(vis,0,sizeof(vis));
   memset(head,-1,sizeof(head));
   cnt=0;
   for(int i=1;i<=m;i++)
   {
       int u,v;
       cin>>u>>v;
       addedg(u,v);
       addedg(v,u);
   }
  // vis[1]=0;
  dfs(1,0);
   if(flag==1)
   {
       cout<<"POSSIBLE";
   }
   else
   {
       cout<<"IMPOSSIBLE";
   }
}