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: Wrong Answer)

#include<stdio.h>
int main()
{	int n,m,i,j,t=0,d=0,e=0,s=0;
int a[200005]={0},b[200005]={0},c[200005],f[200005],g[200005],p[200005],o[200005],u[200005];
	a[-1]=0;
	scanf("%d%d",&n,&m);
	for(i=0;i<m;i++)
	{scanf("%d%d",&a[i],&b[i]);
	}
	if(m==1&&b[0]==n)
	printf("POSSIBLE\n");
	else if(m==1&&b[0]!=n)
	printf("IMPOSSIBLE\n");
	else if(m==2&&(b[0]==n||b[1]==n))
	printf("POSSIBLE\n");
	else if(m==2&&(b[0]!=n||b[1]!=n))
	printf("IMPOSSIBLE\n");
	for(i=0;i<m;i++)
	{
		if(b[i]==n)
		p[e]=a[i],o[e]=b[i-1],u[e]=b[i+1],e++,
		t=1;
		if(a[i]==1)
		c[d]=a[i+1],
		f[d]=a[i-1],
		g[d]=b[i],
		d++;
	}
	if(m>2)
	{
		for(i=0;i<d;i++)
		{
			for(j=0;j<e;j++)
			if(c[i]==p[j]||c[i]==o[j]||f[i]==p[j]||f[i]==o[j]||g[i]==p[j]||g[i]==o[j]||c[i]==u[j]||f[i]==u[j]||g[i]==u[j])
			s=1;
		}
		if(s==1)
		printf("POSSIBLE\n");
		else
		printf("IMPOSSIBLE\n");
	}
}