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;
int N,M;
const int INFTY = 99999; 
vector<list<int>> G;
vector<int> dis;
bool POSSIBLE = false;

void print_list(list<int> list){
  for(auto itr = list.begin(); itr != list.end(); itr++){
    cout << *itr << ' ';
  }
}
void bfs(){
  queue<int> q;
  dis.at(0) = 0;
  q.push(0);
  int u;
  list<int> list;
  while(!q.empty()){
    u = q.front(); q.pop();
    list = G.at(u);
    if(dis.at(u) <= 2 && u == N - 1){
      POSSIBLE = true;
      return;
    }
    for(auto itr = ++list.begin(); itr != list.end(); ++itr){
      if(dis.at(*itr) == INFTY){
        q.push(*itr);
        dis.at(*itr) = dis.at(u) + 1;
      }
    }
  }
}

int main(){
  cin >> N >> M;
  G = vector<list<int>>(N);
  dis = vector<int>(N,INFTY);
  for(int i = 0; i < N; i++){
    G.at(i).push_back(i);
  }
  for(int i = 0; i < M; i++){
    int a,b; cin >> a >> b;
    a--;
    b--;
    G.at(a).push_back(b);
    G.at(b).push_back(a);
  }
  /*for(int i = 0; i < N; i++){
    Print_list(G.at(i));
    cout << endl;
  }*/
  bfs();
  if(POSSIBLE) cout << "POSSIBLE" << endl;
  else cout << "IMPOSSIBLE" << endl;
}