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)

// skyrim belongs to the nords
#include<bits/stdc++.h>
#define mp make_pair
#define pb push_back 
#define sz(a) ((int)(a).size())
#define rep(i, a, b) for(int (i) = (a); (i) < (b); (i)++)
#define dec(i, a, b) for (int (i) = (a); (i) >= (b); (i)--)
#define clr(a,v) memset(a, v, sizeof(a))
#define all(a) (a).begin(),(a).end() 
#define LOGN 18
#define EPS 1e-8
#define fcin ios_base::sync_with_stdio(false)
using namespace std;
typedef long long ll;
typedef double ld;
typedef pair<int,int> pii;
typedef vector<int> vi;

int d[200005], N, M;
vector<int> adj[200005];
vector<int> v;
bool pos = false; 
void dfs(int curr, int k) {
	if (pos) return;
	if (k == 0) {
		if (curr == N) pos = true;
		return;
	}
	// v.push_back(curr);
	for (auto &x: adj[curr]) {
		dfs(x, k - 1);
		if (pos) return;
	}
	// v.pop_back(); 
}
int main(){
	fcin;
	cin.tie(0);
 
	cin >> N >> M;
	int a, b;
	for (int i = 0; i < M; i++) {
		cin >> a >> b;
		adj[a].pb(b);
		adj[b].pb(a);
	} 
	dfs(1, 2);
	if (!pos) cout << "IMPOSSIBLE\n";
	else cout << "POSSIBLE\n";
}