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)

#pragma region _head

#include<iostream>
#include<string>
#include<vector>
#include<algorithm>
#include<cmath>
#include<map>
#include<set>
#include<iomanip>
#include<queue>
#include<stack>
#include<numeric>
#include<utility>
#include<regex>

#pragma region _define

#define int LL
#define f(i,a,b) for(LL i=a;i<b;i++)
#define f_vI(v,n) f(i,0,n)cin>>v[i]
#define f_v2I(v1,v2,n) f(i,0,n)cin>>v1[i]>>v2[i]
#define f_v3I(v1,v2,v3,n) f(i,0,n)cin>>v1[i]>>v2[i]>>v3[i]
#define f_vO(v,n) f(i,0,n)cout<<v[i]<<endl
#define ei else if
#define all(a) a.begin(),a.end()
#define size(s) ((LL)s.size())
#define F first
#define S second
#define check() cout<<"! ! !"
#define endl "\n"
#define _y() cout<<"Yes"<<endl
#define _Y() cout<<"YES"<<endl
#define _n() cout<<"No"<<endl
#define _N() cout<<"NO"<<endl
#define INF 1<<29
#define MOD 1000000007

#pragma endregion

#pragma region _using

using namespace std;

using LL = long long;
using st = string;
using vi = vector<LL>;
using vvi = vector<vi>;
using vvvi = vector<vvi>;
using vc = vector<char>;
using vs = vector<st>;
using vb = vector<bool>;
using vvb = vector<vb>;
using vvvb = vector<vvb>;
using qi = queue<LL>;
using qc = queue<char>;
using qs = queue<st>;
using si = stack<LL>;
using sc = stack<char>;
using ss = stack<st>;
using pi = pair<LL, LL>;
using vpi = vector<pi>;

#pragma endregion

#pragma region _graph

#define node 50  //グラフの最大位数　適宜変更
//bool graph[node][node]; bool visited[node] = { false };
#define link(a,b,m) f(i,0,m){cin>>a[i]>>b[i]; a[i]--; b[i]--; graph[a[i]][b[i]]=graph[b[i]][a[i]]=true;}

#pragma endregion

LL gcd(LL a, LL b) {
	LL r;
	while ((r = a % b) != 0) {
		a = b;
		b = r;
	}
	return b;
}
LL lcm(LL a, LL b) {
	return (a / gcd(a, b) * b);
}

//素数判定
bool is_prime(int n) {
	for (int i = 2; i*i <= n; i++) {
		if (n % i == 0) {
			return false;
		}
	}
	return true;
}

//階乗計算
LL factorial(LL n) {
	LL a = 1, ret = 1;
	while (a < n) {
		a++;
		ret *= a;
		ret %= MOD;
	}
	return ret;
}

#pragma endregion 



/*****************************************************************************/

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

	int n, m;
	cin >> n >> m;
	vi a(m), b(m);
	f_v2I(a, b, m);
	map<int, int>m1;
	f(i, 0, m) {
		m1[a[i]] = b[i];
	}
	f(i, 0, m) {
		if (a[i] == 1 && m1[m1[a[i]]] == n) {
			cout << "POSSIBLE";
			return 0;
		}
	}
	cout << "IMPOSSIBLE";
	return 0;
}