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: Runtime Error)

#include<bits/stdc++.h>
using namespace std;

int main(){
  int n,m;
  cin >> n >> m;
  int a[m] = {0};
  int b[m] = {0};
  int cur = 0,now = 0;
  for(int i = 0;i < m;i++){
    cin >> a[i] >> b[i];
    if(a[i] == 1){
      cur++;
    }
    else if(b[i] == n){
      now++;
    }
  }
  vector<int> start(cur);
  vector<int> goal(now);
  int c = 0,d = 0;
  int k = 0,ok = 0;
  if(cur != 0 && now != 0){
  for(int i = 0;i < m;i++){
    if(a[i] == 1){
      start.at(c) = b[i];
      c++;
    }
    else if(b[i] == n){
      goal.at(d) = a[i];
      d++;
    }
  }
  sort(goal.begin(),goal.end());
  for(int i = 0;i < cur;i++){
    k = lower_bound(goal.begin(),goal.end(),start.at(i)) - goal.begin();
    if(goal.at(k) == start.at(i)){
      ok++;
      break;
    }
  }
  }
  if(ok != 0){
    cout << "POSSIBLE" << endl;
  }
  else{
    cout << "IMPOSSIBLE" << endl;
  }
}