using namespace std;

#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <vector>
#include <string>
#include <algorithm>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <stack>
#include <queue>
#include <utility>
#include <functional>

#define pb push_back
#define f first
#define s second
#define sz size
#define II inline
#define ll long long
#define db double
#define FOR(i,a,b) for(int i=a;i<=b;++i)
#define all(v) v.begin() , v.end()
#define C(v) C.erase( all(v) )
#define FORit(it,v) for(it = (v).begin();it != (v).end();++it)
#define mp make_pair

#define N_MAX 1<<10
#define L_MAX 1<<23
#define oo 1<<28
#define IN "struti.in"
#define OUT "struti.out"

int K(-1),N,M,P;
int S1[N_MAX][N_MAX],S2[N_MAX][N_MAX],A[N_MAX][N_MAX],H[N_MAX][N_MAX];
char buffer[L_MAX];

II int read()
{
	int aux(0);
	for(;buffer[K] > '9' || buffer[K] < '0';++K);
	for(;buffer[K] <= '9' && buffer[K] >= '0';++K)
		aux = aux * 10 + buffer[K] - '0';
	return aux;
}

II void scan()
{
	freopen(IN,"r",stdin);
	freopen(OUT,"w",stdout);
	fread(buffer,1,1<<23,stdin);
	fclose(stdin);
	N = read();
	M = read();
	P = read();
	FOR(i,1,N)
	FOR(j,1,M)
		H[i][j] = read();
}

II void rmq(int X,int Y)
{
	int V1[N_MAX]={0},V2[N_MAX]={0};
	int j,sz1(1),sz2(1);
	FOR(i,1,M)
		memset(A[i],0,sizeof(A[i]));	
	
	FOR(k,1,N)
	{
		V1[0] = V2[0] = V1[1] = V2[1] = sz1 = sz2 = 1;
				
		memset(S1[k],0,sizeof(S1[k]));
		memset(S2[k],1,sizeof(S2[k]));
			
		FOR(i,2,M)
		{
			for(j=V1[0];j >= sz1 && H[i][k] >= H[ V1[j] ][k];--j);
			V1[0] = j;
			V1[ ++V1[0] ] = i;
		
			for(j=V2[0];j >= sz2 && H[i][k] <= H[ V2[j] ][k];--j);
			V2[0] = j;
			V2[++V2[0]] = i;
		
			if(i<=Y)
				continue;
			for(;i-V1[sz1] >= Y+1;++sz1);
			for(;i-V2[sz2] >= Y+1;++sz2);
		
			S1[k][i-Y] = H[ V1[sz1] ][k];
			S2[k][i-Y] = H[ V2[sz2] ][k];
		}	
		
	}
	
	FOR(k,1,M)
	{
		V1[0] = V2[0] = V1[1] = V2[1] = sz1 = sz2 = 1;
		
		FOR(i,2,N)
		{
			for(j=V1[0];j>=sz1 && S1[i][k] >= S1[ V1[j] ][k];--j);
			V1[0] = j;
			V1[ ++V1[0] ] = i;
		
			for(j=V2[0];j>=sz2 && S2[i][k] <= S2[ V2[j] ][k];--j);
			V2[0] = j;
			V2[++V2[0]] = i;
		
			if(i<=X)
				continue;
			for(;i-V1[sz1] >= X+1;++sz1);
			for(;i-V2[sz2] >= X+1;++sz2);
		
			A[i-X][k] = S1[ V1[sz1] ][k] - S2[ V2[sz2] ][k];
		}	
		
	}
	
}

II void solve()
{
	int best(oo),nr(0),X,Y;
	X = read();
	Y = read();
	--X,--Y;
			
	rmq(X,Y);
	FOR(i,1,N-X)
	FOR(j,1,M-Y)	
		if(A[i][j] == best)
			++nr;
		else
			if(A[i][j] < best)
				best = A[i][j],
				nr = 1;
	if(X==Y)
	{
		printf("%d %d\n",best,nr);
		return;
	}	
	rmq(Y,X);
	
	FOR(i,1,N-Y)
	FOR(j,1,M-X)	
		if(A[i][j] == best)
			++nr;
		else
			if(A[i][j] < best)
				best = A[i][j],
				nr = 1;
	printf("%d %d\n",best,nr);
}

int main()
{
	scan();
	for(++P;--P;solve());
	printf("Your Program run for %d ms\n",clock() );
	return 0;
}