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: Accepted)

#include<stdio.h>

int main(void){
    int n,m;
    scanf("%d%d", &n,&m);

    int one_to[n-2];
    int to_n[n-2];
    for(int i=0; i<n-2; i++){
        one_to[i] = 0;
        to_n[i] = 0;
    }

    for(int i=0; i<m; i++){
        int a,b;
        scanf("%d%d", &a, &b);

        if(a==1 && b!=n){
            one_to[b-2] = 1;
        }
        else if(a!= 1 && b==n){
            to_n[a-2] = 1;
        }
    }

    int res = 0;
    for(int i=0; i<n-2; i++){
        if(one_to[i]==1 && to_n[i]==1){
            res = 1;
        }
    }

    if(res){
        printf("POSSIBLE\n");
    }else{
        printf("IMPOSSIBLE\n");
    }

    return 0;
}