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<bits/stdc++.h>
using namespace std;
#define all(vec) vec.begin(),vec.end()
typedef long long int ll;
typedef pair<int,int> P;
const ll MOD=1234567;
const ll INF=1000000010;
const ll LINF=4000000000000000010LL;
const int MAX=200010;
const double EPS=1e-3;
int dx[4]={0,1,0,1};
int dy[4]={0,0,1,1};
struct edge{int to,cost;};
vector<int> dijkstra(int s,vector<edge> g[]){
    vector<int> d(MAX,INF);
    priority_queue<P,vector<P>,greater<P>> q;
    d[s]=0;
    q.push(P(0,s));
    while(!q.empty()){
        P p=q.top();q.pop();
        int v=p.second;
        for(int i=0;i<g[v].size();i++){
            edge e=g[v][i];
            if(d[e.to]>d[v]+e.cost){
                d[e.to]=d[v]+e.cost;
                q.push(P(d[e.to],e.to));
            }
        }
    }
    return d;
}
vector<edge> G[200010];
int main(){
	int n,m;cin>>n>>m;
	for(int i=0;i<m;i++){
		int a,b;cin>>a>>b;a--;b--;
		G[a].push_back({b,1});
		G[b].push_back({a,1});
	}
	vector<int> d=dijkstra(0,G);
	if(d[n-1]==2){
		cout<<"POSSIBLE"<<endl;
	}else{
		cout<<"IMPOSSIBLE"<<endl;
	}
    return 0;
}