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: Time Limit Exceeded)

#include<iostream>
#include<string>
#include<vector>
#include<math.h>
#include<algorithm>
#include<cstdio>
#include<stdio.h>
#include<stack>
#include<sstream>
#include<iomanip>
#include<map>
typedef long long ll;

#define pi 3.14159265358979323846
#define VI vector<int>
#define VLL vector<long long>
#define MAX max_element
#define MIN min_element
#define all(v) v.begin(), v.end()
const ll MOD = 1e9 + 7;
using namespace std;
template<typename T>
T gcd(T a, T b) {
	return b ? gcd(b, a % b) : a;
}
template<typename T>
T lcm(T a, T b) {
	return a * b / gcd(a, b);
}

template<typename T>
T finder(vector<T> vec, T number) {
	auto itr = find(vec.begin(), vec.end(), number);
	size_t index = distance(vec.begin(), itr);
	if (index != vec.size()) { // 発見できたとき
		return 1;
	}
	else { // 発見できなかったとき
		return 0;
	}
}
ll frac(ll n)
{
	if (n == 0) {
		return 1;
	}
	return (n * frac(n - 1)) % MOD;
}

/*char r[4];
int yaju[4];
char op[3];
int sum = 0;
scanf("%s", r);
for (int i = 0;i < 4;i++) {
	yaju[i] = r[i] - '0';
}
*/
int main() {
	ll N, M;cin >> N >> M;
	VLL a(M), b(M);
	bool ok = false;
	ll ans = 0;
	for (int i = 0;i < M;i++) {
		cin >> a[i] >> b[i];
	}
	for (int i = 0;i < M;i++) {
		if (a[i] == 1) {
			ll t = 0;
			while (1) {
				if (b[i] == a[t]) {
					if (b[t] == N) {
						ok = true;
						break;
					}
				}
				if (t == M - 1) break;
				t++;
			}
		}
	}
	if (ok)cout << "POSSIBLE" << endl;
	else cout << "IMPOSSIBLE" << endl;
}