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: Time Limit Exceeded)

#include <bits/stdc++.h>
using namespace std;

#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define REP(i,n)   FOR(i,0,n)
#define ALL(a)     (a).begin(),(a).end()
#define MOD 1000000007

typedef long long int ll;

const ll INF=(ll)1e18;
//const int INF=(1<<30);

void dfs(vector<vector<int>> edge, vector<int> &dist, int n, int d){
  if(d == 3)return;

  dist[n] = d;

  REP(i,edge[n].size()){
    if(dist[edge[n][i]] != -1)continue;
    dfs(edge, dist, edge[n][i], d+1);
  }
}

int main(){
  int N, M;
  cin >> N >> M;

  vector<vector<int>> edge(N+1);

  REP(i,M){
    int a,b;
    cin >> a >> b;
    edge[a].push_back(b);
    edge[b].push_back(a);
  }

  vector<int> dist(N+1,-1);

  dfs(edge, dist, 1, 0);

  // FOR(i,1, dist.size()){
  //   cout << i << " " << dist[i] << endl;
  // }

  cout << (dist[N] != -1  ? "POSSIBLE" : "IMPOSSIBLE") << endl;
}