/*
What does the formula .5rn(3n+1) mean?

r = number of topics
for every topic the score should be at least
.5n(3n+1) -> divide by .5n -> 3n+1
n=2 (1 2 3 4) -> (1 2 3 4) -> min 7 = 3 + 4 = 7
n=4 (1 2 3 4 5 6 7 8 ) -> (.5 1 1.5 2 2.5 3 3.5 4) -> min 13 -> 4 + 3.5 + 3 + 2.5 = 13
Meaning every round we require to have the best matches
So every pair should at least have a score of (n+1 .. 2n) for every topic
The formula enforces that every round we want an optimal matching with max scores

For the first contest divide in 2 groups with scores in (n+1, 2n) and those not
create a vector to if their other scores will provide a winning score
if all vectors have their inverse present in the other group then it is possible, else impossible
By sorting we can make the search doable in O(n log n)
 */


fun main() {
    val (n, r) = readln().split(" ").map { it.toInt() }
    val ranking =
        System.`in`.bufferedReader().lineSequence().take(r).map { line -> line.split(" ").map { it.toInt() } }.toList()
    // build the vectors (as strings for simpler sorting)
    val students = (0 until 2 * n)
        .map { s -> ranking.map { if (it[s] > n) 1 else 0 }.joinToString(separator = "") }
        .groupBy { it[0] }
    // sort for easier search
    val winners = students['1']!!.sorted()
    // flip the inverse of the vectors and sort them
    val losers =
        students['0']!!.map { l -> l.map { if (it == '1') '0' else '1' }.joinToString(separator = "") }.sorted()
    // the matching of winners and the flipped losers should be the same contents
    if (winners.zip(losers).all { (a, b) -> a == b }) println("possible") else println("impossible")
}