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: Time Limit Exceeded)

#include <bits/stdc++.h>
#define rep(i,n) for(int (i)=0;(i)<(n);(i)++)
#define all(x) (x).begin(),(x).end()
#define INF 10000000
using ll = long long ;
ll gcd(ll a, ll b) {
  if(b == 0) return a;
  return gcd(b, a % b);
}
ll lcm(ll a,ll b){
    ll g = gcd(a,b);
    return a / g * b;
}
template< int id > struct CompTupleBy
{
  template<class T> 
  bool operator()( const T& a, const T& b ) const {
    return std::get<id>(a) < std::get<id>(b);
  }
};

using namespace std;

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    int n,m;
    cin >> n >> m;
    vector<int> one;
    vector<int> nvec;
    rep(i,m){
        int a,b;
        cin >> a >> b;
        if (a==1){
            one.push_back(b);
        }else if(b==n){
            nvec.push_back(a);
        }
    }
    for (auto item : one){
        if (find(all(nvec),item) != nvec.end()) {
            cout << "POSSIBLE" << endl;
            return 0;
        }
    }
    cout << "IMPOSSIBLE" << endl;


        
}