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 <vector>
#include <string>
#include <algorithm>
#include <cmath>
#include <queue>
#include <climits>
#include <set>
#include <map>
#include <stack>
#include <iomanip>
#include <tuple>
#define ll long long
using namespace std;
using graph =vector<vector<ll>>;
/*階乗*/
ll factorial(ll n){
    if(n==0){
        return 1;
    }
    else{
        return n*factorial(n-1)%1000000007;
    }
}

/* ユークリッドの互除法 最大公約数*/
/*ll gcd(ll a, ll b) {
	ll r;
	r = a % b;
	while (r != 0) {
		a = b;
		b = r;
		r = a % b;
	}
	return b;
}
*/


/*最小公倍数*/
/*
ll lcm(ll a,ll b){
	return a/gcd(a,b)*b;
}
*/

/*二点間の距離*/
double dist(pair<double,double> a,pair<double,double> b){
	return sqrt(pow((a.first - b.first) ,2) + pow((a.second - b.second),2));
}

//繰り返し自乗法
double ism(double aa,ll p){
    double ap=aa;
    double ans=1;
    while(p>0){
        //cout<<"p="<<p<<",ap="<<ap<<endl;
        if(p&1){//奇数が真
            ans*=ap;
        }
        p/=2;
        ap=ap*ap;
    }
    return ans;

}
//繰り返し自乗法(アマリトリバージョン)
ll ismm(ll aa,ll p,ll m){
    ll ap=aa;
    ll ans=1;
    while(p>0){
        //cout<<"p="<<p<<",ap="<<ap<<endl;
        if(p&1){//奇数が真
            ans=(ans*ap)%m;    
        }
        p/=2;              
        ap=(ap*ap)%m;
    }
    return ans;

}
/*
ll solve(ll i,ll m,vector<ll>t,ll n){
    if(m==0){
        return 1;
    }
    if(i>=n){
        return 0;
    }
    
    
    cout<<i+1<<" "<<m<<endl;
    cout<<i+1<<" "<<m-t[i]<<endl;
    ll res=solve(i+1,m,t,n)||solve(i+1,m-t[i],t,n);
    return res;
}
*/

//小数点12桁
struct all_init
{
    all_init()
    {
        cout << fixed << setprecision(12);
    }
} All_init;


int main()
{
    //頂点sからｔへたどり着けるか
    // 頂点数と辺数
    ll n,m;
    cin>>n>>m;
    graph g(n);
    for(ll i=0;i<m;i++){
        ll a,b;
        cin>>a>>b;
        g[--a].push_back(--b);
        //g[b].push_back(a);
    }
    ll s=0,t=n-1;
    // 頂点 s をスタートとした探索
    vector<ll>dist(n,-1);
    queue<ll>que;
    dist[s]=0,que.push(s);
    while(!que.empty()){
        ll v=que.front();
        que.pop();
        for(auto nv:g[v]){
            if(dist[nv]==-1){
                dist[nv]=dist[v]+1;
                que.push(nv);
            }
        }
    }
     // t に辿り着けるかどうか
    if(dist[t]!=-1&&dist[t]==2){
        cout<<"POSSIBLE"<<endl;
        //cout<<dist[t]<<endl;
    }
    else{
        cout<<"IMPOSSIBLE"<<endl;
        //cout<<dist[t]<<endl;
    }


    

}