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 (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include<stdio.h>
#include<string.h>
#include<math.h>
int s[200000],e[200000];
int main()
{
    int b[200000][2],n,i,m,ans,j,k,t;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        for(i=0;i<m;i++)
            scanf("%d%d",&b[i][0],&b[i][1]);
        if(m<2)
            printf("IMPOSSIBLE\n");
        else
        {
            for(i=0,j=0,k=0,ans=0;i<m;i++)
            {
                if(b[i][1]==n)
                {
                    e[k]=b[i][0];
                    k++;
                }
                if(b[i][0]==1)
                {
                    s[j]=b[i][1];
                    j++;
                }
            }
            for(i=0;i<j;i++)
            {
                if(s[i+1]!=s[i])
                {
                    for(t=0;t<k;t++)
                        if(s[i]==e[t])
                        {
                            ans=1;
                            break;
                        }
                }
            }
            if(ans)
                printf("POSSIBLE\n");
            else
                printf("IMPOSSIBLE\n");
        }
    }
    return 0;
}