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>

#define rep2(x,fr,to) for(int (x)=(fr);(x)<(to);(x)++)
#define rep(x,to) for(int (x)=0;(x)<(to);(x)++)
#define repr(x,fr,to) for(int (x)=(fr);(x)>=(to);(x)--)
#define all(c) (c).begin(),(c).end()
#define sz(v) (int)(v).size()

using namespace std;
typedef long long ll;  typedef pair<int,ll> pii;
typedef vector<int> VI; typedef vector<ll> VL;
const int MD = (int)1e9 + 7;
void dbg(){ cerr << "\n"; }  template <typename T,typename ...T2>
void dbg(const T& fst, const T2&...rst){ cerr << fst << ": "; dbg(rst...); }


int main(){
	//cin.tie(0); ios_base::sync_with_stdio(false);
	int n, m;
	cin >>n >>m;
	vector<VI> r(n);
	
	rep(i, m){
		int a,b;
		cin >>a >>b;
		r[a-1].push_back(b-1);
		r[b-1].push_back(a-1);
	}

	const int INF =1<<29;
	VI ds(n, INF); ds[0] = 0;
	queue<int> q; q.push(0);
	
	for(; !q.empty(); ){
		int v = q.front(); q.pop();
		for(auto x: r[v]){
			if(ds[x] == INF){
				ds[x] = ds[v] + 1;
				q.push(x);
			}
		}
	}
	
	cout <<(ds[n-1]<3? "POSSIBLE": "IMPOSSIBLE") <<endl;
	
	return 0;
}