// Solution to Orienteering
// Author: Thomas Beuman

// Time complexity: O((n-1)!)
// Memory: O(n^2 + (n/2)!)

// Expected result: accepted

#include <algorithm>
#include <cstdio>
using namespace std;

const int nmax = 14;

long long Dis[nmax][nmax];

long long Trip1[720], Trip2[720]; // array sizes: (ceil(nmax/2-1))!

// Considers all permutations of points in subset s as intermediate stops between 0 and p;
// stores distances in Trip[] and returns number of permutations
int tryallperms (int n, int p, int s, long long Trip[])
{	// Base case
	if (s == 0)
	{	Trip[0] = Dis[0][p];
		return 1;
	}
	int m = 0, i, k = 0;
	int Point[nmax/2];
	long long d = 0;
	// Dissect subset
	for (i = 1; i < n; i++)
		if (s & (1<<i))
			Point[k++] = i;
	do
	{	d = Dis[0][Point[0]];
		for (i = 1; i < k; i++)
			d += Dis[Point[i-1]][Point[i]];
		d += Dis[Point[k-1]][p];
		Trip[m++] = d;
	}
	while (next_permutation(Point, Point+k));
	return m;
}

int main()
{	int n, i, j, m1, m2, p, q, s;
	long long L;
	scanf("%d %lld", &n, &L);
	for (i = 0; i < n; i++)
		for (j = 0; j < n; j++)
			scanf("%lld", &Dis[i][j]);
	// 0 = start point, p = mid point
	for (p = 1; p < n; p++)
	{ q = (p==1 ? 2 : 1); // Arbitrary other point
		// Try all subsets of size n/2-1 that do not involve 0, p or q
		for (s = 0; s < (1<<n); s += 2)
			if (!(s & (1<<p)) && !(s & (1<<q)) && __builtin_popcount(s) == n/2-1)
			{	m1 = tryallperms(n, p, s, Trip1);
				// Invert subset (ignoring 0 and p)
				m2 = tryallperms(n, p, s ^ ((1<<n)-2) ^ (1<<p), Trip2);
				// See if two trip lengths combined give L
				for (i = 0; i < m1; i++)
					for (j = 0; j < m2; j++)
						if (Trip1[i] + Trip2[j] == L)
						{	printf("possible\n");
							return 0;
						}
			}
	}
	printf("impossible\n");
	return 0;
}