#include <iostream>
using namespace std;
const int N = 300;
int a[N][N];//存值
int b[N][N];//判断某个点是否出现过
int n,k;
string path;
int dx[] = {-1,-1,0,1,1,1,0,-1};
int dy[] = {0,1,1,1,0,-1,-1,-1};
bool dfs(int x,int y,int cur,int pos)
{if(pos == n*n && x == n-1 && y== n-1){cout << path << endl;return true;}for(int i = 0;i<=7;i++){int bx = x+dx[i];int by = y+dy[i];if(b[bx][by]) continue;if(bx<0 || bx>n-1 || by<0 || by>n-1)continue;if(((cur == k-1) && (a[bx][by] != 0)) || (cur < k-1 && a[bx][by] != cur+1)) continue;if(i==1&&b[bx][by-1]&&b[bx+1][by]) continue;else if(i==3&&b[bx-1][by]&&b[bx][by-1]) continue;else if(i==5&&b[bx-1][by]&&b[bx][by+1]) continue;else if(i==7&&b[bx+1][by]&&b[bx][by+1]) continue;b[bx][by] = 1;path+=to_string(i);if(dfs(bx,by,a[bx][by],pos+1))return true;b[bx][by] = 0;path.pop_back();}return false;}
int main()
{cin >> n >> k;for(int i = 0;i<n;i++){for(int j = 0;j<n;j++){cin >> a[i][j];} }b[0][0] = 1;if(!dfs(0,0,a[0][0],1))cout << -1 << endl; return 0;
}
