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<iostream>
#include<stdio.h>
#include<vector>
#include<algorithm>
#include<set>
#include<string>
#include<map>
#include<string.h>
#include<complex>
#include<math.h>
#include<queue>
#include <functional>
#include<time.h>
#include <stack>
#include<iomanip>
using namespace std;
#define rep(i,a,n) for(llint i=(a);i<(n);i++)
#define ll long long
#define llint long long int
#define sort(s) sort(s.begin(),s.end())
#define reverse(v)  reverse(v.begin(), v.end());
#define Yes(ans) if(ans)cout<<"Yes"<<endl; else cout<<"No"<<endl;
#define YES(ans) if(ans)cout<<"YES"<<endl; else cout<<"NO"<<endl;
#define hei(a) vector<a>
#define whei(a) vector<vector<int>>
#define UF  UnionFind
constexpr auto mod = 1000000007;
//辞書順はnext_permutation( begin( v ), end( v ) );やで！


//2のn乗を求めるよ！！！
int ni(int n) {
	if (n == 0)return 1;
	int x = ni(n / 2);
	x *= x;
	if (n % 2 == 1)x *= 2;
	return x;
}

//フィボナッチ数列のx番目を求めるよ！
llint f(int x, vector<llint> &s) {
	if (x == 0)return 0;
	if (x == 1)return 1;
	if (s[x] != 0)return s[x];
	return  s[x] = f(x - 1, s) + f(x - 2, s);
	
}

//aとｂの最大公約数を求めるよ！
llint gcd(llint a, llint b) {
	if (a < b)swap(a, b);
	if (b == 0)return a;
	return gcd(b, a % b);
}



// a^n mod を計算する
long long modpow(long long a, long long n, long long mod) {
	long long res = 1;
	while (n > 0) {
		if (n & 1) res = res * a % mod;
		a = a * a % mod;
		n >>= 1;
	}
	return res;
}

// mod. m での a の逆元 a^{-1} を計算するよ！
long long modinv(long long a, long long m) {
	long long b = m, u = 1, v = 0;
	while (b) {
		long long t = a / b;
		a -= t * b; swap(a, b);
		u -= t * v; swap(u, v);
	}
	u %= m;
	if (u < 0) u += m;
	return u;
}


//aCbをmod.mで割った余りを求める
ll int c(ll int a, ll int b, ll int m) {
	ll int ans = 1;
	for (ll int i = 0; i < b; i++) {
		ans *= a - i;
		ans %= m;
	}
	for (ll int i = 1; i <= b; i++) {
		ans *= modinv(i, m);
		ans %= m;
	}
	return ans;
}

//m進数のn桁を全列挙やで
void dfs(int m, int n,hei(int) a) {
	stack<string> st;
	st.push("");
	while (!st.empty()) {
		string now = st.top();
		st.pop();
		if (now.size() == n) cout << now << endl;
		else {
			for (int i = m - 1; i >= 0; i--) {
				string next = now + to_string(i);
				st.push(next);
			}
		}
	}

}
void press(vector<int>& v) { v.erase(unique((v).begin(), (v).end()), v.end()); }

llint min(llint a, llint b) {
	if (a < b)return a;
	return b;
}
llint max(llint a, llint b) {
	if (a < b)swap(a, b);
	return a;
}
llint p(int n) {
	if(n==1)return 1;
	return n * p(n - 1);
}
struct UnionFind {
	vector<int> par; // par[i]:iの親の番号　(例) par[3] = 2 : 3の親が2

	UnionFind(int N) : par(N) { //最初は全てが根であるとして初期化
		for (int i = 0; i < N; i++) par[i] = i;
	}
	int size(int a) {
		return par[root(a)];
	}

	int root(int x) { // データxが属する木の根を再帰で得る：root(x) = {xの木の根}
		if (par[x] == x) return x;
		return par[x] = root(par[x]);
	}

	void unite(int x, int y) { // xとyの木を併合
		int rx = root(x); //xの根をrx
		int ry = root(y); //yの根をry
		if (rx == ry) return; //xとyの根が同じ(=同じ木にある)時はそのまま
		par[rx] = ry; //xとyの根が同じでない(=同じ木にない)時：xの根rxをyの根ryにつける
	}

	bool same(int x, int y) { // 2つのデータx, yが属する木が同じならtrueを返す
		int rx = root(x);
		int ry = root(y);
		return rx == ry;
	}
};

int main() {
	int n, m;
	cin >> n >> m;
	UF tree(n+1);
	rep(i, 0, m) {
		int a, b;
		cin >>a >> b;
		tree.unite(a-1, b-1);
                tree.unite(b-1,a-1);
	}
	if (tree.same(0, n-1) == true&&tree.size(0)==2)cout << "POSSIBLE" << endl;
	else cout << "IMPOSSIBLE" << endl;
	return 0;
}