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 Python (3.4.3) (judge verdict: Wrong Answer)

n, m = [int(x) for x in input().split()]
isl = []
mark = [False] * n
l = [0] * n
for i in range(n):
    isl.append([])
for i in range(m):
    a, b = [int(x) for x in input().split()]
    isl[a - 1].append(b - 1)
    isl[b - 1].append(a - 1)
 
q = []
q.append(0)
found = False
while len(q):
    v = q.pop()
    mark[v] = True
    for child in isl[v]:
        if not mark[child]:
            l[child] = l[v] + 1
            q.append(child)
print("POSSIBLE" if (l[n - 1] == 2 or l[n - 1] == 1) else "IMPOSSIBLE")