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 (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include<stdio.h>

#define M 300000

int main(){
  int n, m, i, j, sec1[M], t, fir_n[M], sec_n[M];
  int ans=0, f, s, k=0, l=0;
  
  scanf("%d%d", &n, &m);

  for(i=0; i<m; i++){
    scanf("%d%d", &f, &s);
    if(f==1){
      sec1[k]=s;
      k++;
    }else{
      fir_n[l]=f;
      sec_n[l]=s;
      l++;
    }
  }

  for(i=0; i<k; i++){
    for(j=0; j<l; j++){
      if(sec1[i]==fir_n[j] && sec_n[j]==n){
	ans=1;
	break;
      }
    }
    if(ans==1){
      break;
    }
  }
  
  if(ans==1){
    printf("POSSIBLE\n");
  }else{
    printf("IMPOSSIBLE\n");
  }
  
  return 0;
}