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 "math.h"
#include<vector>
#include<queue>	
using namespace std;
vector<vector<int>> edges;
vector<int> mark;
vector<int> d;
queue<int> q;
int n,m;
void BFS(int start){
	q.push(start);
	d[start]=0;
	mark[start]=1;
	while(!q.empty()){
		int v=q.front();
		q.pop();
		for(int i=0;i<edges[v].size();i++){
			if(mark[edges[v][i]]!=1){
				d[edges[v][i]]=d[v]+1;
				mark[edges[v][i]]=1;
				q.push(edges[v][i]);
			}

		}

	}
}
using namespace std;
int main(){
	cin>>n>>m;
	edges.resize(n);
	mark.resize(n);
	d.resize(n);
	for(int i=0;i<n;i++){
		d[i]=4;
	}
	int p,q;
	for(int i=0;i<m;i++){
		cin>>p>>q;
		p--;
		q--;
		edges[p].push_back(q);
		edges[q].push_back(p);
	}
	BFS(0);
	
	if(d[n-1]<3){
		cout<<"POSSIBLE";
	}
	else{
		cout<<"IMPOSSIBLE";
	}
	cin>>d[n-1];
}