#include <bits/stdc++.h>
#define soclose 1e-8
using namespace std;

struct pt {
	double x,y,z;
	};

double getY(double X, pt &A, pt &B) {
	return A.y-(A.x-X)*(A.y-B.y)/(A.x-B.x);}

double getX(double Y, pt &A, pt &B) {
	return A.x-(A.y-Y)*(A.x-B.x)/(A.y-B.y);}

double getZx(double X, pt &A, pt &B) {
	return A.z+(X-A.x)*(A.z-B.z)/(A.x-B.x);}

double getZy(double Y, pt &A, pt &B) {
	return A.z+(Y-A.y)*(A.z-B.z)/(A.y-B.y);}

int main() {
	int T;
	cin >> T;
	for(int t =0; t < T; t++) {
		int R,C;
		cin >> R >> C;
		vector< vector<int> > H(R,vector<int>(C));
		for(int i =0; i < R*C; i++) cin >> H[i/C][i%C];
		vector< vector<bool> > F(R,vector<bool>(C,false));
		pair<int,int> P[2];
		for(int k =0; k < 2; k++) {
			cin >> P[k].first >> P[k].second;
			P[k].first -=1;
			P[k].second -=1;}
		if(P[0].first == P[1].first && P[0].second == P[1].second) {
			cout << "The shortest path is 0 steps long.\n";
			continue;}
		
		for(int i =0; i < R; i++) for(int j =0; j < C; j++) for(int k =0; k < 2; k++) {
			if(P[k].first == i && P[k].second == j) {F[i][j] =true; continue;}
			bool ok =true;
			pt A; A.x =i+0.5; A.y =j+0.5, A.z =H[i][j]+0.5;
			pt B; B.x =P[k].first+0.5; B.y =P[k].second+0.5, B.z =H[P[k].first][P[k].second]+0.5;
			if(P[k].first != i) {
				for(int x =min(i,P[k].first)+1; x < max(i,P[k].first); x++) {
					double y =getY(x,A,B);
					double z =getZx(x,A,B);
					int Y =(int)(y+soclose);
					if(abs(y-round(y)) < soclose) continue;
					if(H[x+1][Y]-soclose >= z) ok =false;
					if(H[x][Y]-soclose >= z) ok =false;}
				}
			if(P[k].second != j) {
				for(int y =min(j,P[k].second)+1; y < max(j,P[k].second); y++) {
					double x =getX(y,A,B);
					double z =getZy(y,A,B);
					int X =(int)(x+soclose);
					if(abs(x-round(x)) < soclose) continue;
					if(H[X][y+1]-soclose >= z) ok =false;
					if(H[X][y]-soclose >= z) ok =false;}
				}
			if(ok) F[i][j] =true;}
		
		queue< pair<int,int> > q;
		int dx[] ={0,0,1,-1};
		int dy[] ={1,-1,0,0};
		q.push(P[0]);
		vector< vector<int> > D(R,vector<int>(C,R*C+42));
		D[P[0].first][P[0].second] =0;
		while(!q.empty()) {
			pair<int,int> p =q.front();
			for(int k =0; k < 4; k++) if(min(p.first+dx[k],p.second+dy[k]) >= 0 && p.first+dx[k] < R && p.second+dy[k] < C) 
				if(F[p.first+dx[k]][p.second+dy[k]] && H[p.first][p.second]+1 >= H[p.first+dx[k]][p.second+dy[k]] && H[p.first][p.second]-3 <= H[p.first+dx[k]][p.second+dy[k]] && D[p.first+dx[k]][p.second+dy[k]] > D[p.first][p.second]+1) {
				D[p.first+dx[k]][p.second+dy[k]] =D[p.first][p.second]+1;
				q.push(make_pair(p.first+dx[k],p.second+dy[k]));}
			q.pop();}
		
		if(D[P[1].first][P[1].second] > R*C) cout << "Mission impossible!\n";
		else cout << "The shortest path is " << D[P[1].first][P[1].second] << " steps long.\n";}
	}