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: Wrong Answer)

#include <iostream>
#include <vector>
#include <set>
#include <algorithm>
#include <iterator>
using namespace std;
typedef long long ll;
ll N, M;

#define rep(i, n) for(int i = 0; i < (int)(n); i++)
#define rep_s(i, start, n) for(int i = start; i < (int)(n); i++)
int main(void){
	cin >> N >> M;
  	ll a_num, b_num;
  	vector<double> a, b;
  //cout << "a" << endl;
  	rep(i, M){
      //cout << "a" << endl;
      cin >> a_num >> b_num;
      //cout << "cin finished" << endl;
      if (a_num == 1){
        a.push_back(b_num);
      }
      else if (b_num == N){
        b.push_back(a_num);
      }
      //cout << "done" << endl;
    }
  
  	vector<double> v;
    set_intersection(a.begin(), a.end(), b.begin(), b.end(), back_inserter(v));
  	if (!v.empty()){
      cout << "POSSIBLE" << endl;
    }
  else {
    cout << "IMPOSSIBLE" << endl;
  }
  return 0;
}