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 <vector>
#include <set>
#include <algorithm>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
#define for1(i,n) for (ll i=0;i<(n);i++)
#define for2(i,m,n) for (ll i=(m);i<(n);i++)
#define for3(i,m,n,d) for (ll i=(m);i<(n);i+=(d))
#define DEBUG 0
template <class T>
void dumplist(T list)
{
    for (auto item : list) {
        cout << item << " ";
    }
    cout << endl;
}

string solve()
{
    ll N,M,a,b;
    cin >> N >> M;
    set<ll> s1;
    set<ll> s2;
    set<ll> s3;

    for1(i,M) {
        cin >> a >> b;
        if (a == 1) {
            s1.insert(b);
        }
        if (b == N) {
            s2.insert(a);
        }
    }
    set_intersection(s1.begin(),s1.end(),s2.begin(),s2.end(),inserter(s3,s3.end()));
    //dumplist(s1);
    //dumplist(s2);
    //dumplist(s3);
    //cout << s3.size() << endl;
    if (s3.size() > 0) {
        return "POSSIBLE";
    }
    return "IMPOSSIBLE";
}

int main()
{
    do {
        cout << solve() << endl;
    } while (DEBUG);
}