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;

#define REP(i, n)   for(int i = 0; i < (n); i++)
#define REPS(i, n)  for(int i = 1; i <= (n); i++)
#define RREP(i, n)  for(int i = (n)-1; i >= 0; i--)
#define RREPS(i, n) for(int i = (n); i > 0; i--)
#define ALL(v) v.begin(), v.end()
#define RALL(v) v.rbegin(), v.rend()
#define UNIQUE(v) v.erase(unique(v.begin(), v.end()), v.end())
#define mp make_pair
#define mt make_tuple
#define pb push_back
#define YES cout << "YES" << endl
#define Yes cout << "Yes" << endl
#define yes cout << "yes" << endl
#define NO cout << "NO" << endl
#define No cout << "No" << endl
#define no cout << "no" << endl

using ll  = long long;
using pi  = pair<int, int>;
using pl  = pair<ll, ll>;
using vi  = vector<int>;
using vl  = vector<ll>;
using vs  = vector<string>;
using vb  = vector<bool>;
using vvi = vector<vi>;
using vvl = vector<vl>;

const int MOD = 1e9 + 7;
//const int MOD = 998244353;
const int INF = 1e9 + 7;
const ll  INFL = 1e18;
const double PI = 3.141592653589793;
const double EPS = 1e-9;

template<class T> bool chmax(T &a, const T &b) { if(a < b) { a = b; return true; } return false; }
template<class T> bool chmin(T &a, const T &b) { if(a > b) { a = b; return true; } return false; }

signed main()
{
    cin.tie(0);
    ios::sync_with_stdio(false);
    cout << fixed << setprecision(20);
    
    int N, M; cin >> N >> M;
    vi e(N), a(M), b(M); 
    REP(i, M) cin >> a[i] >> b[i], a[i]--, b[i]--;
    REP(i, M)
    {
        if(a[i] == 0) e[b[i]] = 1;
        if(b[i] == 0) e[a[i]] = 1;
    }
    REP(i, M)
    {
        if((a[i] == N-1 && e[b[i]] == 1) || b[i] == N-1 && e[a[i]] == 1)
        {
            cout << "POSSIBLE" << endl;
            return 0;
        }
    }
    cout << "IMPOSSIBLE" << endl;
}