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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
 
#define int long long
#define pb push_back
#define ub upper_bound
#define lb upper_bound
#define mp make_pair
#define F first
#define S second
#define FOR(i,a,b) for(int (i)=(a);(i)<(b);(i)++)
#define REP(i,n) FOR(i,0,n)
#define ALL(a) (a).begin(),(a).end()
#define RALL(a) (a).rbegin(),(a).rend()
#define SORT(c) sort((c).begin(),(c).end())
#define ve vector
#define vi vector<int>
#define vp vector<pair<int,int>>
#define vvi vector<vector<int>>
 
using ll = long long;
const ll INF = LLONG_MAX - 100;
ll mod = 1e9 + 7;
int dx[] = {-1,0,1,0}, dy[] = {0,1,0,-1};
vector<ll> prime;

class fact {
public:
	int fmod = 1e9+7;
	vector<int> fac, inv;
	fact (int n, int Mod = 1e9+7) {
		fmod = Mod;
		fac = vector<int>(n,0);
		inv = vector<int>(n,0);
		fac[0] = 1; for (int i = 1; i < n; i++) fac[i] = fac[i-1] * i % fmod;
		for (int i = 0;i < n;i++) inv[i] = fact::POW(fac[i],fmod-2);
	}
	ll nCr(ll n, ll r) {return (fac[n] * inv[r] % fmod) * inv[n-r] % fmod;}
	ll POW(ll a, ll b) {ll c = 1; while (b > 0) {if (b & 1) {c = a * c%fmod;}a = a * a%fmod; b >>= 1;}return c;}
};
 
template <class T = ll> T in() {T x; cin >> x; return (x);}
void DEBUG(vector<int> a) {for(int i=0;i<a.size();i++)cout<<a[i]<<" ";cout<<endl;}
void EMP(int x) {cout<<"!!!"<<x<<"!!!"<<endl;}
ll GCD(ll a, ll b) {ll c; while (b != 0) {c = a % b; a = b; b = c;}return a;}
ll LCM(ll a, ll b) {return a * b / GCD(a, b);}
ll POW(ll a, ll b) {ll c = 1; while (b > 0) {if (b & 1) {c = a * c%mod;}a = a * a%mod; b >>= 1;}return c;}
void PRI(ll n) {bool a[n + 1LL]; for (int i = 0; i < n + 1LL; i++) {a[i] = 1LL;}for (int i = 2; i < n + 1LL; i++) {if (a[i]) {prime.pb(i); ll b = i; while (b <= n) {a[b] = 0; b += i;}}}}
template <typename T> T chmin(T& a, T b) {if(a>b)a=b;return a;}
template <typename T> T chmax(T& a, T b) {if(a<b)a=b;return b;}
bool isSqrt(ll a) {return pow(sqrt(a),2) == a ? 1 : 0;}
void YesNo(bool a) {if (a) cout << "Yes"; else cout << "No"; cout << endl;}
void yesno(bool a) {if (a) cout << "yes"; else cout << "no"; cout << endl;}
void YESNO(bool a) {if (a) cout << "YES"; else cout << "NO"; cout << endl;}

class UnionFind {
private:
	vector<int> par;
public:
	UnionFind(int N) { par = vector<int>(N, -1); }
	int find(int x);
	ll size(int x);
	void unite(int x, int y);
	bool same(int x, int y);
};

int UnionFind::find(int x) {
	if (par[x] < 0) return x;
	else return par[x] = find(par[x]);
}
 
ll UnionFind::size(int x) {
	return -par[find(x)];
}
 
void UnionFind::unite(int x, int y) {
	x = find(x);
	y = find(y);
 	if (x == y) return;
	//大きい方に小さい方をくっ付ける
	if (size(x) < size(y)) swap(x, y);
	par[x] += par[y];
	par[y] = x;
}
 
bool UnionFind::same(int x, int y) {
	x = find(x);
	y = find(y);
	return x == y;
}

bool solve() {
    int n,m; cin >> n >> m;
    UnionFind uf(n);
    REP (i,m) {
        int a,b; cin >> a >> b; --a, --b;
        uf.unite(a,b);
    }
    bool ok = false;
    FOR (i,1,n-1) {
        if ((uf.same(i,0) && uf.same(i,n-1))) {
            ok = true;
        }
    }
    if (ok) cout << "POSSIBLE" << endl;
    else cout << "IMPOSSIBLE" << endl;
}

signed main() {
	cin.tie(0);
	ios::sync_with_stdio(false);
	solve();
}