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;


#define file "file"
#define mp make_pair
#define pb push_back
typedef long long ll;
typedef complex<double> point;
typedef pair<int, int> pii;
typedef vector<int> vi;

#define DEBUG 0
#define dout if(DEBUG) cout

const int MAXN = 2e5 + 5;

vi g[MAXN];
int cnt[MAXN];

void update(int v){
    for(int i = 0; i < (int)g[v].size(); i++){
        int to = g[v][i];
        cnt[to]++;
    }
}

void solve(){
    int n, m;
    cin >> n >> m;
    int v, to;
    for(int i = 0; i < m; i++){
        scanf("%d%d", &v, &to);
        g[v].pb(to);
        g[to].pb(v);
    }
    update(1);
    update(n);
    for(int i = 2; i < n; i++){
        if(cnt[i] == 2){
            cout << "POSSIBLE";
            return;
        }
    }
    cout << "IMPOSSIBLE";
}

int main() {
    #ifdef NASTYA
        assert(freopen("input.txt", "r", stdin)); assert(freopen("output.txt", "w", stdout));
    #else
    // assert(freopen(file".in", "r", stdin)); assert(freopen(file".out", "w", stdout));
    #endif
    int t = 1;
    while(t--) {
        solve();
    }
    return 0;
}