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>
#define ALL(a) (a).begin(), (a).end()
#define RALL(a) (a).rbegin(), (a).rend()
#define pb push_back
#define FOR(i, a, b) for (int i = (a); i < (b); ++i)
#define rep(i, n) FOR(i, 0, n)
#define ll long long
using namespace std;

const ll P = 1000000007;
int gcd(int a, int b) { return b != 0 ? gcd(b, a % b) : a; }
int lcm(int a, int b) { return a / gcd(a, b) * b; }

int main()
{
  cout << fixed << setprecision(10);
  int N,M;
  cin >> N >> M;
  vector<int> a(M),b(M);
  set<int> s;
  rep(i,M) 
  {
    cin >> a[i] >> b[i];
    a[i] --;
    b[i] --;
    if(a[i]==0)
    {
      s.insert(b[i]);
    }
  }
  rep(i,M)
  {
    if(b[i]==N-1&&s.count(a[i]))
    {
      cout << "POSSIBLE" << endl;
      return 0;
    }
  }
  cout << "IMPOSSIBLE" << endl;
  return 0;
}