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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<cstdio>
#include<cstring>
const int MAXN=200000+10;

struct edge{
    int to,next;
}edges[MAXN];
int heads[MAXN];
int tot;

void addedge(int a,int b)
{
    edges[tot].to=b;
    edges[tot].next=heads[a];
    heads[a]=tot++;
}

int main()
{
    int n,m;
    while(scanf("%d%d",&n,&m)==2)
    {
        tot=0;
        memset(heads,-1,sizeof(heads));
        for(int i=0;i<m;i++)
        {
            int a,b;
            scanf("%d%d",&a,&b);
            addedge(a,b);
        }

        bool has=false;
        for(int i=heads[1];i!=-1;i=edges[i].next)
        {
            for(int j=heads[edges[i].to];j!=-1;j=edges[j].next)
            {
                if(edges[j].to==n){printf("POSSIBLE\n");has=true;break;}
            }
            if(has)break;
        }
        if(has==false)printf("IMPOSSIBLE\n");
    }
    return 0;
}