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)

#include <iostream>
#include <vector>
#include <numeric>
#include <algorithm>
#include <cstring>
#include <map>
#include <set>
#include <cmath>
#define FOR(i, a, b) for (int i = (a); i < (b); ++i)
#define REP(i, n) for (int i = 0, i##_len = (n); i < i##_len; ++i)
#define INF 1145141919
typedef long long int ll;
using namespace std;

template <typename T>
istream &operator>>(istream &is, vector<T> &vec)
{
    for (T &x : vec)
        is >> x;
    return is;
}

ll gcd(ll m, ll n)
{
    if ((0 == m) || (0 == n))
        return 0;
    while (m != n)
    {
        if (m > n)
            m = m - n;
        else
            n = n - m;
    }
    return m;
}

ll lcm(ll m, ll n)
{
    if ((0 == m) || (0 == n))
        return 0;
    return ((m / gcd(m, n)) * n);
}

int main(){
    int n,m;
    cin >> n >> m;

    vector<int> a(m), b(m);
    REP(i,m){
        cin >> a[i] >> b[i];
    }
    
    int now=1,cnt=1;

    while(now!=n and cnt!=2){
        auto result = find(a.begin(), a.end(),now);
        now = b[*result];
        cnt++;
    }

    if(cnt==2 and now==n){
        cout << "POSSIBLE" << endl;
    }else{
        cout << "IMPOSSIBLE" << endl;
    }

    return 0;
}