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: Runtime Error)

#include <stdio.h>

int main(){
	long int N, M;
	scanf("%d %d",&N, &M);

	long int a[N], b[N];
	int flag[2][N];
	int k = 0;
	for(int i = 0; i < N; i++){
		flag[0][i] = 0;
		flag[1][i] = 0;
	}
	for(int i = 0; i < M; i++){
		long long int buf1, buf2;
		scanf("%d %d", &buf1, &buf2);
		if(buf1 == 1){
			a[k] = buf1;
			b[k] = buf2;
			flag[0][b[k]-1] = 1;
			k++;
		} else if(buf2 == N) {
			a[k] = buf1;
			b[k] = buf2;
			flag[1][a[k]-1] = 1;
			k++;
		}
	}

	for(int i = 0; i < M; i++){
		if(flag[0][i] == 1 && flag[1][i] == 1){
			printf("POSSIBLE\n");
			return 0;
		}
	}

	printf("IMPOSSIBLE\n");

	return 0;
}