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 <iostream>
#include <string>
#include <vector>
#include <algorithm>
#include <cmath>
#include <set>
#include <map>
#include <queue>
#include<limits.h>
#define rep(i, n) for(int i = 0; i < n; i++)
#define REP(i, m, n) for(int i = m; i < n; i++)
#define ALL(v) v.begin(),v.end()
#define RALL(v) v.rbegin(),v.rend()
#define check(v) rep(i,v.size()) cout << v[i] << " ";\
cout << endl
#define INF 1e8
typedef long long ll;
using namespace std;
//オーバーフローに気をつけろよおおおおおお
//確認忘れるなよおおおおおお
const int N=2e5+5;
int cnt[N],n,m;
int main()
{
	cin>>n>>m;
	for(int i=0;i<m;i++)
	{
		int a,b;
		cin>>a>>b;
		if(a==1||a==n)cnt[b]++;
		if(b==1||b==n)cnt[a]++;
	}
	for(int i=1;i<=n;i++)
	{
		if(cnt[i]>=2)
		{
			cout<<"POSSIBLE\n";
			return 0;
		}
	}
	cout<<"IMPOSSIBLE\n";

    return 0;
}