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)

/**
 *      ABC 68 C
 *      author  : kyomukyomupurin
 *      created : 2018-09-11 23:14:13
**/

#include <bits/stdc++.h>
typedef long long ll;
using namespace std;

int main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    
    int N, M; cin >> N >> M;
    int a[M] = {}; int b[M] = {};
    for (int i = 0; i < M; ++i) {
        cin >> a[i] >> b[i];
    }

    int step1[M] = {}; int step2[M] = {};
    for (int i = 0; i < M; ++i) {
        if (a[i] == 1){
            step1[i] = b[i];
        }
        else if (b[i] == N){
            step2[i] = a[i]; 
        }
    }

    //int size1 = sizeof(step1) / sizeof(*step1); int size2 = sizeof(step2) / sizeof(*step2);

    for (int i = 0; i < M; ++i) {
        for (int j = 0; j < M; ++j) {
            if (step1[i] == step2[j] && step1[i] * step2[j] != 0){
                cout << "POSSIBLE" << '\n'; return 0;
            }            
        }        
    }
    cout << "IMPOSSIBLE" << '\n';

    return 0;
}