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>
#include<math.h>

int main()
{
  int n,m;
  
  scanf("%d%d",&n,&m);
  
  int a[m+1],b[m+1];
  int x[m+1];
  int i,j=1;
  
  a[0]=0,b[0]=0;
  
  for(i=0;i<m+1;i++)
  {
    x[i]=0;
  }
  
  for(i=1;i<=m;i++)
  {
    scanf("%d %d",&a[i],&b[i]);
  }
  
  for(i=1;i<=m;i++)
  {
    if(a[i]==1)
    {
      x[j]=b[i];
      j++;
    }
  }
  
  for(i=1;i<=m;i++)
  {
    for(j=1;x[j]!=0;j++)
    {
     if(a[i]==x[j])
     {
      if(b[i]==n)
      {
        printf("POSSIBLE\n");
        return 0;
      }
     }
    }
  }
    
  printf("IMPOSSIBLE\n");
  
  
  return 0;
}