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)

// lcmとかはg++ -std=c++17 default.cppみたいなかんじで

#include <bits/stdc++.h>
#define mod 1000000007
#define INF 1001001001
#define ll long long
#define ln cout<<endl
#define Yes cout<<"Yes"<<endl
#define No cout<<"No"<<endl
#define double long double
#define REP(i,m,n) for(ll i=(ll)(m);i<(ll)(n);i++)
#define rep(i,n) REP(i,0,n)
#define rep1(i,n) REP(i,1,n+1)
using namespace std;

int main() {
    cin.tie(0);
   	ios::sync_with_stdio(false);

    ll a,b,c,d,m,n,maxi=0,f=0,mini=INF,sum=0;
    string str;
    set<ll> start;
    set<ll> end;

    //vector<vector<ll>> v(tate,vector<ll> (yoko));
    cin>>n>>m;
    rep(i,m){
        cin>>a>>b;
        if(a==1) start.insert(b);
        else if(b==n) end.insert(a); 
    }
    for(auto itr=start.begin();itr!=start.end();++itr){
        if(end.find(*itr)!=end.end()){
            cout<<"POSSIBLE"<<endl;
            return 0;

        }
    }
    cout<<"IMPOSSIBLE"<<endl;


    return 0;
}