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 <iostream>
#include <string>
#include <utility>
#include <algorithm>
#include <map>
#include <vector>
#include <cmath>
#include <cstdlib>
#include <queue>
#include <stack>

using namespace std;

#define REP(i, n) for(int i = 0;i < n;i++)
#define REPR(i, n) for(int i = n;i >= 0;i--)
#define FOR(i, m, n) for(int i = m;i < n;i++)
#define FORR(i, m, n) for(int i = n;i >= m;i--)
#define REPO(i, n) for(int i = 1;i <= n;i++)
#define ll long long
#define INF 1999999999
#define MINF -1999999999
#define INF64 5999999999999999999
#define ALL(n) n.begin(),n.end()
#define ABS(a,b) max(a,b) - min(a,b)

bool s[300000];
int main() {
	int n, m;
	bool ans = false;
	cin >> n >> m;
	REP(i, m) {
		int a, b;
		cin >> a >> b;
		if (a == 1) {
			if (s[b])ans = true;
			else s[b] = true;
		}
		if (b == n) {
			if (s[a])ans = true;
			else s[a] = true;
		}
	}
	cout << (ans ? "POSSIBLE" : "IMPOSSIBLE") << endl;
}