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# (Mono 4.6.2.0) (judge verdict: Time Limit Exceeded)

using System;
using System.Text;
using System.Collections.Generic;
using System.Linq;
class Solve
{
    class Pair
    {
        public int Key { get; }
        public int Value { get; }
        public Pair(int a, int b)
        {
            Key = a;
            Value = b;
        }
    }

    public static int Main()
    {
        var nm = Console.ReadLine().Split().Select(int.Parse).ToArray();
        int n = nm[0], m = nm[1];
        var pairs = new Pair[m];
        for (int i = 0; i < m; i++)
        {
            var ab = Console.ReadLine().Split().Select(int.Parse).ToArray();
            pairs[i] = new Pair(ab[0], ab[1]);
        }

        var secondWays = pairs.Where(e => e.Value == n).OrderBy(e => e.Key).ToArray();

        if (!secondWays.Any())
        {
            Console.WriteLine("IMPOSSIBLE");
            return 0;
        }
        var lastWay = secondWays[secondWays.Length - 1];
        var beginWay = secondWays[0];
        var firstWays = pairs.Where(e => e.Key == 1).Where(e => e.Value >= beginWay.Key).Where(e => e.Value <= lastWay.Key)/*.OrderBy(e => e.Value)*/.ToArray();

        foreach (var firstWay in firstWays)
        {
            var way = secondWays.FirstOrDefault(e => e.Key == firstWay.Value);
            if (way != null)
            {
                Console.WriteLine("POSSIBLE");
                return 0;
            }
        }

        Console.WriteLine("IMPOSSIBLE");

        return 0;
    }
}