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;
typedef long long ll;
#define FOR(i,n) for(int i=0;i<(n);i++)
#define MOD 1000000007
#define all(x) (x).begin(),(x).end()
template<class T> inline bool chmax(T& a,T b){if(a<b){a=b;return 1;}return 0;}
template<class T> inline bool chmin(T& a,T b){if(a>b){a=b;return 1;}return 0;}
template<class T> inline T GCD(T a,T b){T c;while(b!=0){c=a%b;a=b;b=c;}return a;}
template<class T> inline T LCM(T a,T b){return a/GCD(a,b)*b;}

int main(){
	vector<ll> g[200010];
	ll n,m;
	cin >> n >> m;
	ll a[200010],b[200010];
	FOR(i,m){
		cin >> a[i] >> b[i];
		g[a[i]].push_back(b[i]);
		g[b[i]].push_back(a[i]);
	}
	vector<ll> v;

	FOR(i,g[1].size()){
		v.push_back(g[1][i]);
	}
	vector<ll> h;
	FOR(j,g[n].size()){
		h.push_back(g[n][j]);
	}
	h.push_back(0);
	h.push_back(1000000000000);
	sort(all(h));
	FOR(i,v.size()){
		auto itr=lower_bound(all(h),v[i]);
		if(*itr==v[i]){
			cout << "POSSIBLE" << endl;
			return 0;
		}
	}

	cout << "IMPOSSIBLE" << endl;


}