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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ull = unsigned long long;
using P = pair<int, int>;
const ll INF = 1e16;

void solver(int N, int M, vector<vector<int>> &ab, bool &flg){

    queue<int> q;
    vector<int> cost(N+1, -1);

    q.push(1);
    cost.at(1) = 0;

    //  bfs
    while( !q.empty() ){
        int buf;
        buf = q.front();
        q.pop();
//        cout << "buf = "s << buf << endl;

        for( auto np: ab.at(buf) ){
           if( cost.at(np) != -1 ){
               continue;
           }
           cost.at(np) = cost.at(buf) + 1;
//           cout << np << " "s << cost.at(np) << " "s << cost.at(buf) << endl;
           q.push(np);
        }
    }
    if( cost.at(N) == 2){
        flg = true;
    }

//    for( int i = 0; i < N+1; ++i){
//        cout << "cost = "s << cost.at(i) << endl;
//    }

    return;
}

int main(){

    int N, M;
    cin >> N >> M;
    vector<vector<int>> ab(N+1);
    for( int i = 0; i < M; ++i){
        int a, b;
        cin >> a >> b;
        ab.at(a).push_back(b);
        ab.at(b).push_back(a);
    }

    bool flg = false;
    solver(N, M, ab, flg);

    if( flg ){
        cout << "POSSIBLE"s << endl;
    }
    else{
        cout << "IMPOSSIBLE"s << endl;
    }
    return 0;
}