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 <stdlib.h>


int main()
{

        int n, m;
        //int *a, *b;
        //int *c, *d;
        int num1 = 0;
        int num2 = 0;

        scanf("%d %d", &n, &m);

        int a[200000];
        int b[200000];
        int c[200000];
        int d[200000];


        /*
        a = (int *)malloc(sizeof(int)*m);
        b = (int *)malloc(sizeof(int)*m);
        c = (int *)malloc(sizeof(int)*m);
        d = (int *)malloc(sizeof(int)*m);

        */

        for (int i=0; i<m; i++) {
                scanf("%d %d", &a[i], &b[i]);
                if (a[i] == 1) {
                        c[num1] = i;
                        num1++;
                }
                if (b[i] == n) {
                        d[num2] = i;
                        num2++;
                }

        }

        for (int i=0; i<num1; i++) {
                //search start with b[i]
                for (int j=0; j<num2; j++)
                        if (a[d[j]] == b[c[i]]) {
                                // if next is N
                                if (b[d[j]] == n) {
                                        printf("POSSIBLE\n");
                                        return 0;
                                }
                        }

        }

        printf("IMPOSSIBLE\n");

        return 0;
}