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>
int main(void) {
  int a,b,i,j,k,tag = 0,aa,bb;
  int ax[200001],bx[200001];
  scanf("%d %d",&a,&b);
  for (i = 0; i < b; ++i) {
    scanf("%d %d",&ax[i],&bx[i]);
  }
    for (j = 0; j < b; ++j) {
      if (ax[j] == 1) {
        for (k = 0; k < b; ++k) {
          if (bx[j] == ax[k] && bx[k] == a) {
            tag = 1;
            break;
          } else if (bx[j] == bx[k] && ax[k] == a) {
            tag = i;
            break;
          }
        }
      } else if (bx[j] == 1) {
        for (k = 0; k < b; ++k) {
          if (ax[j] == bx[k] && ax[k] == a) {
            tag = 1;
            break;
          } else if (ax[j] == ax[k] && bx[k] == a) {
            tag = i;
            break;
          }
        }
      }
    }
  
 if (tag == 1) {
   printf("%s","POSSIBLE");
 } else {
   printf("%s","IMPOSSIBLE");
 }
            return 0;
            }