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 <iostream>
#include <vector>
#include <utility>
using namespace std;
    
int n, m;
vector<pair<int, int>> v;
vector<int> w;

void center_point(){
    for(auto x: v){
        if(n == x.first)
            w.push_back(x.second);
        if(n == x.second)
            w.push_back(x.first);
    }
}

bool search(){
    for(auto x: w){
        for(auto y: v){
            auto a=y.first; 
            auto b=y.second;
            if(a==x && b==1) return true;
            if(b==x && a==1) return true;
        }
    }
    return false;
}

void solve(){
    center_point(); // Nから逆にたどって行ける島のリストをwに突っ込む
    if(search()){
        cout << "POSSIBLE" << endl;
    }else{
        cout << "IMPOSSIBLE" << endl;
    }
}

int main(){
    cin >> n >> m;
    v.reserve(m);
    w.reserve(m);
    int a, b;
    for(int i=0; i<m; i++){
        cin >> a >> b;
        v.push_back(make_pair(a, b));
    }
    solve();
}