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 <bits/stdc++.h>

#define F first
#define S second
#define llong long long
#define ullong unsigned long long
#define mp make_pair
#define pb push_back
#define sz(v) (int)v.size()

using namespace std;

const int MXN = (int)1e6 + 10;
const int INF = (int)1e9 + 7;
const llong LINF = (llong)1e18 + 10;
const double EPS = (double)1e-9;
const double PI = (double)acos(-1.0);

bool b1[MXN], b2[MXN];
int n, m;

int main(){
	#ifdef LOCAL
		freopen("input.txt", "r", stdin);
	#endif
	scanf("%d%d", &n, &m);
  for(int i = 1; i <= m; ++i){
  	int u, v;
  	scanf("%d%d", &u, &v);
  	if(u > v)
  		swap(u, v);
  	if(u == 1)
  		b1[v] = true;
  	if(v == n)
  		b2[u] = true;	
  }
  for(int i = 2; i < n; ++i){
  	if(b1[i] == b2[i] && b1[i] == true){
  		printf("POSSIBLE");
  		return 0;
  	}
  }
  printf("IMPOSSIBLE");
  return 0;
}