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;

typedef long long ll;
typedef long double ld;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef pair<int, int> ii;
typedef vector<ii> vii;

#define fi first
#define se second
#define endl '\n'
#define MOD(n,k) ( (n % k) + k ) % k
#define isOn(S, j) (S & (1 << j) )
#define setBit(S, j) (S | (1 << j ) )
#define LSB(S) (S & (-S))

int main () {
	ios_base::sync_with_stdio(0); cin.tie(0);
	
	int n, m, x, y;
	vii a;

	cin >> n >> m;

	bitset<200005> bs;

	for (int i = 0; i < m; i++) {
		cin >> x >> y;
		a.emplace_back(x, y);
		if (x == 1)
			bs[y] = 1;
	}

	for (int i = 0; i < m; i++) {
		if (bs[a[i].fi] && a[i].se == n) {
			cout << "POSSIBLE";
			return 0;
		}
	}

	cout << "IMPOSSIBLE";

	return 0;
}