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: Wrong Answer)

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<cctype>
#include<cmath>
#include<cstdlib>
#include<queue>
#include<ctime>
using namespace std;
int m,n,q;
struct node
{  int a,b;}s[200001];
int fa[200001];
int find(int k)
{    q++;
     if(fa[k]==k)return k;
       else {return find(fa[k]);}}
int main()
{     int i,j,k;q=-1;
      scanf("%d%d",&n,&m);
      for(i=1;i<=m;i++)
      {  scanf("%d%d",&s[i].a,&s[i].b);}
      for(i=1;i<=n;i++)fa[i]=i;
      for(i=1;i<=m;i++)
         fa[s[i].b]=s[i].a;
      if(find(n)==1)cout<<"POSSIBLE"<<endl;
        else cout<<"IMPOSSIBLE"<<endl;
      /*for(i=1;i<=m;i++)
         if(s[i].a==1){q++;
		   if(find(s[i].b)){cout<<"POSSIBLE"<<endl;return 0;}}
		 else break;
	  cout<<"IMPOSSIBLE"<<endl;*/
	  return 0;
}