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>
using namespace std;

using ll = long long;
#define countof(a) (sizeof(a)/sizeof((a)[0]))
#define rep(i, n) for(int i = 0; i < (n); i++)
#define rep1(i, n) for(int i = 1; i < (n); i++)
#define all(a) begin(a),end(a)
template<class T> istream& operator >> (istream& s, vector<T>& v)
{ for (T& x: v) { s >> x; } return s;}
template<class T> void cout_v(const vector<T>& c)
{ for (int i = 0; i < c.size(); i++) cout << c[i] << endl; }
template<class T> void cout_h(const vector<T>& c)
{ for (int i = 0; i < c.size(); i++) cout << c[i] << (i==c.size()-1?"\n":" "); }
template<class T> inline bool chmin(T& a, T b) {return a>b?a=b,true:false;}
template<class T> inline bool chmax(T& a, T b) {return a<b?a=b,true:false;}
void Yes(bool f){cout<<(f?"Yes":"No")<<endl;}
#define SP << " " <<
#define FMT(a) #a<<":"<<a 
#define FMT2(a,b) #a<<":"<<a<<", "<<#b<<":"<<b 
#define FMT3(a,b,c) #a<<":"<<a<<", "<<#b<<":"<<b<<", "<<#c<<":"<<c 
#define FMT4(a,b,c,d) #a<<":"<<a<<", "<<#b<<":"<<b<<", "<<#c<<":"<<c<<", "<<#d<<":"<<d 

int main()
{
    cin.tie(0); ios::sync_with_stdio(false);
    //cout<<std::fixed<<std::setprecision(10);

    int N, M; cin >> N >> M;
    vector<int> f(N+1, 0);
    
    int a, b;
	bool possible = false;
    rep(i, M) {
        cin >> a >> b;
		if ((a == 1 && (f[b] |= 1) == 3)
		 || (b == N && (f[a] |= 2) == 3)
		) {
			possible = true;
			break;
        }
    }
    cout << (possible ? "POSSIBLE" : "IMPOSSIBLE") << endl;
}