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 REP(i,n) for (int i = 0; i < (n); i++)
#define RREP(i, s, n) for (int i = s; i < (n); i++)
#define ALL(a) a.begin(), a.end()
#define RALL(a) a.rbegin(), a.rend()
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }
using namespace std;
using ll = long long;
typedef pair<int, int> pint;
typedef pair<ll, ll> pll;
const ll MOD = 1000000007;
const ll INF = MOD * MOD;
const int inf = (1<<29);

int main() {

    ll N, M;
    cin >> N >> M;
    vector<vector<ll>> G(N);
    for (ll i = 0; i < M; ++i) {
        ll a, b;
        cin >> a >> b;
        a--; b--;
        G[a].push_back(b);
        G[b].push_back(a);
    }

    vector<ll> dist(N, -1); // 全頂点を「未訪問」に初期化
    queue<ll> que;

    // 初期条件 (頂点 0 を初期ノードとする)
    dist[0] = 0;
    que.push(0);

    // BFS 開始 (キューが空になるまで探索を行う)
    while (!que.empty()) {
        ll v = que.front(); // キューから先頭頂点を取り出す
        que.pop();

        // v から辿れる頂点をすべて調べる
        for (ll nv : G[v]) {
            if (dist[nv] != -1) continue; // すでに発見済みの頂点は探索しない

            // 新たな白色頂点 nv について距離情報を更新してキューに追加する
            dist[nv] = dist[v] + 1;
            que.push(nv);
        }
    }

    cerr << dist[N-1] << endl;

    if (dist[N-1] <= 2 && dist[N-1] != -1) {
        cout << "POSSIBLE" << endl;
    } else {
        cout << "IMPOSSIBLE" << endl;
    }


    return 0;
}